H C Verma solutions, ELECTRIC FIELD AND POTENTIAL, Chapter-29, Concepts of Physics, Part-II

Electric Field and Potential

Questions for Short Answer

   1. The charge on a proton is +1.6x10⁻¹⁹ C and that on an electron is -1.6x10⁻¹⁹ C. Does it mean that the electron has a charge 3.2x10⁻¹⁹ C less than the charge of a proton? 

Answer: No. Here, it means that the charge on a proton is equal to that on an electron, but the nature of the charge is opposite.   

    2. Is there any lower limit to the electric force between two particles placed at a separation of 1 cm?  

Answer: The electric force between two particles is given as Kq₁q₂/r². K is a constant, and r = 1 cm is also a constant in this case; hence, the lowest limit of the electric force will be when q₁ and q₂ are lowest. Since the lowest possible charge on a particle will be equal to that on one electron or one proton i.e. = e. Hence, the lower limit of the electric force between two particles =Ke²/r². Attraction or repulsion. 

    3. Consider two particles A and B having equal charges and placed at some distance. The particle A is slightly displaced towards B. Does the force on B increase as soon as particle A is displaced? Does the force on particle A increase as soon as it is displaced?  

Answer: The interaction between electrically charged particles is due to the electric field around them. These fields can travel at the speed of light. Since particle A can not be displaced more quickly than the speed of light, the answers to both questions are yes. 

    4. Can a gravitational field be added vectorially to an electric field to get a total field?  

Answer: No. Both types of fields are totally different in nature and do not interact with each other.   

    5. Why does a phonograph record attract dust particles just after it is cleaned? 

Answer: When the record is cleaned, it gets electrically charged due to rubbing. Hence, it attracts dust particles by induction or those particles that have an opposite charge from the record.    

    6. Does the force on a charge due to another charge depend on the charges present nearby?  

Answer: No. The force on a charge due to another charge is fully independent of the charges present nearby.   

    7. In some old texts, it is mentioned that 4π lines of force originate from each unit positive charge. Comment on the statement in view of the fact that 4π is not an integer. 

Answer: The total solid angle around a point is 4π steradians. Though 4π is not an integer, it signifies the all-around space about a unit positive charge where these lines of force are equispaced. 

    8. Can two equipotential surfaces cut each other?   

Answer: An equipotential surface is the surface on which each point has the same potential. Two equipotential surfaces will be for two different potentials. They can not cut each other because if they do, there will be two different potentials at the points where they cut, which will be outright wrong.  

    9. If a charge is placed at rest in an electric field, will its path be along a line of force? Discuss the situation when the line of forces is straight and when they are curved. 

Answer: It will depend upon the nature of the line of force, whether it is straight or curved. When the charge is placed at rest in the electric field, the force on the charge is along the tangent drawn on the line of force at that point. When the charge is released, it starts to move along the direction of the force.

     Now, for the straight line of force, the movement of the charge and the force are always along the straight line. Thus, its path will be along the line of force. See the point P in the diagram below.

Diagram for Q-9

 

   For the curved line of force, the charge starts to move along the force, the direction of which is tangential to the line of force (At point A). This force will accelerate it, and the charge will very soon get some velocity. At this point, the charge will have a force along the tangent (Force F at point B), and due to the curvature, this direction of the force is not the same direction as when the charge was released. Also, now there is another force of inertia due to the motion (Force F') acting on it, and its direction will be along the velocity attained. So the charge will move along the resultant (Force R) of the two forces and not along the line of the electric force.     

    10. Consider the situation shown in Figure (29-Q1). What are the signs of q₁ and q₂? If the lines are drawn in proportion to the charge, what is the ratio q₁/q₂? 

Figure for Q-10

Answer: Since the lines are entering q₁ it is a negative charge. The lines are emerging out of q₂ hence it is a positive charge. 

    If the lines are drawn proportional to the charge, we see 6 lines entering q₁ and 18 lines emerging out of q₂. Thus the ratio,

q₁/q₂ = 6/18 = 1:3.

 

    11. A point charge is taken from point A to point B in an electric field. Does the work done by the electric field depend on the path of the charge?  

Answer: No, the work done by the electric field depends only on the potential difference between points A and B. It is independent of the path of the charge.   

    12. It is said that the separation between the two charges forming an electric dipole should be small. The small compared to what?  

Answer: The separation between the two charges forming an electric dipole should be small compared to the distance of the point from the center of the dipole, where we intend to find the field due to the dipole.   

    13. The number of electrons in an insulator is of the same order as the number of electrons in a conductor. What is then the basic difference between a conductor and an insulator? 

Answer: The basic difference between a conductor and an insulator is that the electrons can move freely in a conductor, but they can not move over long distances in an insulator due to more rigid bonding in the molecules.  

    14. When a charged comb is brought near a small piece of paper, it attracts the piece. Does the paper become charged when the comb is brought near it?  

Answer: No, the paper does not get charged; only the electrons in the paper are rearranged. The concentration of opposite charge to the comb becomes higher near the end, which is why it is attracted towards the comb. The net charge on the paper remains the same. 

OBJECTIVE-I

   1. Figure (29-Q2) shows some of the electric field lines corresponding to an electric field. The figure suggests that

Figure for Q-1

(a) E_A > E_B > E_C

(b) E_A = E_B = E_C

(c) E_A = E_C > E_B

(d) E_A = E_C < E_B 

Answer: (c).  

Explanation: There are an equal number of electric field lines within the same areas at A and C, but fewer lines at B. 

   

    2. When the separation between two charges is increased, the electric potential energy of the charges

(a) increases

(b) decreases

(c) remains the same

(d) may increase or decrease   

Answer: (d).  

Explanation: Electric potential energy is defined as the negative of the work done by the electric forces. If both the charges are of the same nature, the work done by the forces is positive, and the potential energy is negative on increasing the separation. But it will be the opposite if the charges are of a different nature. 

  

    3. If a positive charge is shifted from a low potential region to a high potential region, the electric potential energy

(a) increases

(b) decreases

(c) remains the same

(d) may increase or decrease.    

Answer: (a). 

Explanation: In shifting a positive charge from a low potential energy region to a high potential energy region, work will be done against the electric force, i.e., the work done by the electric force is negative, so the potential energy is positive, i.e. the electric potential energy increases. 

 

    4. Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential

(a) continuously increases

(b) continuously decreases

(c) increases then decreases

(d) decreases, then increases.    

Answer: (d).  

Explanation: Suppose two positive charges q and q are placed at a distance r from each other. Potential at a point P at a distance x from A (x<r) due to these charges will be

V = q/4πєₒx + q/4πєₒ(r-x)

   =(q/4πєₒ){(r-x+x)/x(r-x)}

   =qr/4πєₒx(r-x)

Differentiating V with respect to x, we get

dV/dx =(qr/4πєₒ)[-1/x²(r-x) +1/x(r-x)²]

  ={qr/4πєₒx(r-x)}[1/(r-x)-1/x]

  =qr(2x-r)/4πєₒx²(r-x)²

At x → 0, dV/dx →Infinity(-ve)

at x =r/2, dV/dx = 0, 

at x →r, dV/dx →Infinity(+ve)

So the potential first decreases, and after x =r/2 it increases.   

 

    5. The electric field at the origin is along the positive X-axis. A small circle is drawn with the center at the origin, cutting the axes at points A, B, C, and D having coordinates (a, 0), (0, a), (-a, 0), (0,-a) respectively. Out of the points on the periphery of the circle, the potential is minimum at

(a) A

(b) B

(c) C

(d) D.    

Answer: (a).  

Explanation: Since the potential decreases at the maximum rate along the electric field, the minimum potential among these four will be minimum at the point on the x-axis, i.e., at A.

  

    6. If a body is charged by rubbing it, its weight

(a) remains precisely constant

(b) increases slightly

(c) decreases slightly

(d) may increase slightly or may decrease slightly.     

Answer: (d).  

Explanation: When a body is charged by rubbing, either it gets some electrons from another body to be negatively charged or loses some electrons to be positively charged. So either its weight may increase slightly or decrease slightly due to the electron transfers.

    

    7. An electric dipole is placed in a uniform electric field. The net electric force on the dipole 

(a) is always zero

(b) depends on the orientation of the dipole

(c) can never be zero

(d) depends on the strength of the dipole.   

Answer: (a).  

Explanation: In a uniform electric field, both ends will experience equal and opposite force (because the ends of a diapole have equal and opposite charge), hence the net force on the diapole will always be zero. (But there will be the torque).

    

    8. Consider the situation of the figure (29-Q3). The work done in taking a point charge from P to A is W_A, from P to B is W_B, and from P to C is W_C.

(a) W_A <W_B < W_C

(b) W_A > W_B > W_C

(c) W_A = W_B =W_C

(d) None of these   

Figure for Q-8

Answer: (c).  

Explanation: The three points A, B, and C are at equal distances from q. So the potential at each of the three points will be equal. Thus, the work done in bringing a point charge from P to these points will be equal. 

   

    9. A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is

(a) zero

(b) positive

(c) negative

(d) zero if the charge Q is at the center and nonzero otherwise.   

Answer: (a).  

Explanation: The work done by the electric field depends upon the final and initial positions, not on the path followed by the charge. In one complete revolution, the rotating charge comes to the same point; hence, no work is done.   

OBJECTIVE-II

   1. Mark out the correct options.

(a) The total charge of the universe is constant.

(b) The total positive charge of the universe is constant

(c) The total negative charge of the universe is constant.

(d) The total number of charged particles in the universe is constant.  

Answer: (a).    

Explanation: The sum of all charges with the proper sign is always constant in the universe. It can be understood with an example of a neutral atom. The charge is zero. If an electron jumps over to another atom/molecule, it will get one unit of positive charge while the receiving atom will get one negative charge, thus the total charge of these two atoms will still be zero.

 

   2. A point charge is brought in an electric field. The electric field at a nearby point

(a) will increase if the charge is positive

(b) will decrease if the charge is negative

(c) may decrease if the charge is positive

(d) may decrease if the charge is negative.  

Answer: (c), (d). 

Explanation: The field around a nearby point charge will not be the same when it is placed in an electric field, whether it is positive or negative. It may increase or decrease depending on the position of the selection of a nearby point because the electric field will be the vector summation of the fields of the point charge and the original field.     

 

   3. The electric field and the electric potential at a point are E and V, respectively.

(a) If E = 0, V must be zero.

(b) If V = 0, E must be zero.

(c) If E ≠ 0, V cannot be zero.

(d) If V ≠ 0, E cannot be zero. 

Answer: None.    

Explanation: E is a vector, and V is a scalar. At a point, E may be zero in a direction, but it may not be zero in the other direction. The direction of E is not clearly stated. Only at a very, very far away from a charge, E may be zero in all directions, and V will be zero here. So it is not necessary that V =0 if E =0 and vice versa(Option (a) and (b) not true). Also, E is zero along a direction where V is constant (option (d) is wrong).

   We know that the potential V at a point r (vector) where the electric field is E, is given as

V = ∫E.dr 

Thus, even if E is not zero but the angle between E and r is 90°, V will be zero. So option (c) is not true. 

 

   4. The electric potential decreases uniformly from 120 V to 80 V as one moves on the x-axis from x =-1 cm to x =+1 cm. The  electric field at the origin

(a) must be equal to 20 V/cm

(b) maybe equal to 20 V/cm

(c) maybe greater than 20 V/cm

(d) maybe less than 20 V/cm.  

Answer: (b),  (c).    

Explanation: We know,

-(dV/dr) = E.Cosθ

→E = -(1/cosθ)(dV/dr)

    =-(1/cosθ)*(-40/2)

    =20/Cosθ

Since the value of cosine is from 0 to 1, the value of E at the origin may be from 20 and above, depending upon the angle between the x-axis and the direction of vector E. The direction of E is along the maximum rate of change of V.

 

   5. Which of the following quantities do not depend on the choice of zero potential or zero potential energy?

(a) the potential at a point

(b) the potential difference between two points

(c) the potential energy of a two-charge system

(d) change in the potential energy of a two-charge system.     

Answer: (b), (d).    

Explanation: The potential at a point and the potential energy of a two-particle system are defined with respect to a reference point, which is the zero potential or zero potential energy. The difference in potential or change in potential energy does not depend upon it because it is with respect to each other.

 

   6. An electric dipole is placed in an electric field generated by a point charge.

(a) The net electric force on the dipole must be zero.

(b) The net electric force on the dipole may be zero.

(c) The torque on the dipole due to the field must be zero.

(d) The torque on the dipole due to the field may be zero. 

Answer: (d).    

Explanation: The arm of a dipole is assumed to be much much less than other dimensions, hence the field around it may be assumed to be parallel. Thus, there will be equal and opposite electric forces at its two ends, and the net force on the dipole will be zero. Since the forces are equal and opposite, there will be a torque acting on the dipole in this electric field, the magnitude of which will depend on the orientation of the dipole. If the length of the dipole is along the electric field, the torque will be zero due to the zero arm length of the couple of electric forces.

 

   7. A proton and an electron are placed in a uniform electric field.

(a) The electric forces acting on them will be equal.

(b) The magnitude of the forces will be equal.

(c) Their accelerations will be equal.

(d) The magnitudes of their accelerations will be equal.

 

Answer: (b).    

Explanation: The electric force F on a charge q placed in an electric field E is given as,

F = qE, 

Since the magnitude of q is equal on a proton and on an electron, but it is positive on the proton and negative on the electron, the magnitude of F will be the same on both of them, but the direction will be opposite. Hence, option (b) is correct.

       The acceleration is given as

a = F/m. The mass m of an electron is much less than that of a proton; hence, the magnitude will not be equal, and the direction will be opposite.

 

   8. The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero.

(a) It is uniform in the region.

(b) It is proportional to r

(c) It is proportional to r²

(d) It increases as one goes away from the origin. 

Answer: (c).    

Explanation: Let the Electric field E = kr. Where r is a constant. Since the difference in the potential is given as

Vr - Vo = ∫Edr, the limit of integration is from the origin to a distance r. And given Vo =0. Thus,

Vr = ∫krdr =kr²/2

so, Vr ∝ r². Option (c) is correct.     

EXERCISES

   1. Find the dimensional formula of εₒ. 

Answer: From Coulomb's law,

F =(1/4πεₒ)*(q₁q₂/r²)

→εₒ = (1/4πF)(q₁q₂/r²)

4π is a number and dimensionless. The dimensional formula of F =[MLT⁻²]

for q₁ and q₂ =[IT]

for r = [L]

Putting in the formula,

Dimension of εₒ = [IT][IT]/[MLT⁻²][L]²

   =[I²M⁻¹L⁻³T⁴] 

 

   2. A charge of 1.0 C is placed at the top of your college building, and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times of your weight is this force?  

Answer: q₁ =q₂ =1.0 C

r =2.0 km =2000 m

The force exerted by the charges on each other,

F =(1/4πεₒ)*(q₁q₂/r²)

  =9x10⁹*(1.0*1.0/2000²)

  =2.25x10³ N (repulsive)

If the weight of a student is assumed to be 50 kg =500 N (approx), then the above force is 2.25x10³/500 =4.5 times the weight of the student. This shows that a 1.0 C charge is quite large compared to the daily life occurrences of charges.  

  

   3. At what separation should two equal charges, 1.0 C each, be placed so that the force between them equals the weight of a 50 kg person? 

Answer: Force between the charges, F =weight of the person, W =mg

=50*9.8 N =490 N

We have to find out the separation between the charges =r =?

F = 9x10⁹*1.0*1.0/r²

→r² =9x10⁹/490 =0.184x10⁸

→r =0.43x10⁴ =4.3x10³ m .   

  

   4. Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person? 

Answer: F = mg =50*9.8 N =490 N

r = 1.0 m, q₁ =q₂ =q =?

F =9x10⁹(q₁*q₂/r²)

490 =9x10⁹*q²/1.0²

→q² =49/9x10⁸

→q =√5.44/10⁴ 

   =2.3x10⁻⁴ C.   

  

   5. Find the electric force between two protons separated by a distance of 1 fermi (1 fermi = 10⁻¹⁵ m). The protons in a nucleus remain at a separation of this order.

Answer: r =1.0x10⁻¹⁵ m,

Charge on each proton, q =1.6x10⁻¹⁹ C

Hence, the force between the protons,

F =9x10⁹*(1.6x10⁻¹⁹)²/(1.0x10⁻¹⁵)² N

 =9*2.56x10 N 

 =230.4 N.   

  

   6. Two charges 2.0x10⁻⁶ C and 1.0x10⁻⁶ C are placed at a separation of 10 cm. Where should be a third charge be placed such that it experiences no net force due to these charges?

Answer: Such a place will be at a point where the electric field by each charge is equal but opposite in direction. The direction of the electric field will be opposite each other on the line joining them. Let the magnitude of the electric fields at point P at a distance x cm from the larger charge is equal.

Diagram for Q-6

 Electric field by larger charge = Electric field by smaller charge at point P

→kQ/x² = kq/r²

→r²/x² = q/Q   {r =10-x cm}

→ (10-x)²/x² =1/2

→(100-20x+x²)/x² =1/2

→x² =200-40x+2x²  

→x²-40x+200 =0

→(x²-2.20.x+20²)-200 =0

→(x-20)² =200

→x-20 =14.14 or -14.14

→x =34.14 or 5.86 cm

x can not be equal to 34.14 cm because it is >10 cm, where the direction of the electric fields will not be the opposite. Hence, the third charge should be placed at 5.86 cm from the larger charge, in between the two charges.  

   

   7. Suppose the second charge in the previous problem is -1.0x10⁻⁶ C. Locate the position where a third charge will not experience a net force. 

Answer: In this case, the direction of the electric fields will be the same between the charges on the line joining them. The direction of the fields will be opposite only outside this 10 cm segment on the extended line. As we have solved the x =34.14 cm or 5.86 from the larger charge in the previous problem, here x cannot be equal to 5.86 cm. So the third charge will not experience a net force at 34.14 cm from the larger charge on the line joining the charges, on the side of the smaller charge.  

  

   8. Two charged particles are placed at a distance of 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge? 

Answer: The magnitude of the electric force F is proportional to q'q"/r². Since r is fixed = 1.0 cm, hence F will be minimum when q' and q" are both minima. The minimum possible charge is that of an electron = 1.6x10⁻¹⁹ C.

Now F =9x10⁹*(1.6x10⁻¹⁹)²/(1/100)² N

 =9*2.56x10⁻²⁵ N

=23x10⁻²⁵ N

=2.3x10⁻²⁴ N.   

 

 

   9. Estimate the number of electrons in 100 g of water. How much is the total negative charge on those electrons? 

Answer: One molecule of water contains one oxygen atom and two hydrogen atoms. A hydrogen atom has one electron, while an oxygen atom has eight electrons. So one molecule of water contains 10 electrons. 

      Molecular weight of water =2+16 =18 units. So 100 g of water is equal to 100/18 =5.56 gram moles of water. Since one gram mole contains Avagadro's Number of molecules, the total number of molecules in the given quantity of water =5.56*6.023x10²³

=3.35x10²⁴ 

Hence, the total number of electrons in 100 g of water =3.35x10²⁴*10

=3.35x10²⁵

The negative charge on one electron =1.602x10⁻¹⁹ C

Hence total negative charge on the electrons of 100 g water,

  =3.35x10²⁵*1.602x10⁻¹⁹ C

  =5.36x10⁶ C.

   10. Suppose all the electrons of 100 g of water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positively charged particle. If these two particles are placed on 10.0 cm away from each other, find the force of attraction between them. Compare it with your weight.  

Answer: Separation between the charges, r = 10.0 cm =0.10 m.

Since water molecules are neutral, the total positive charge on all the nuclei lumped together will be equal to the total negative charge on all the electrons. 

Now, q₁ =q₂ =5.36x10⁶ C.

Force of attraction between two charges, 

F =(1/4πεₒ)*(q₁q₂/r²)

 =9x10⁹*(5.36x10⁶)²/(0.10)² N

 =258x10²³ N

 =2.58x10²⁵ N.  

Assuming self-weight = 50 kg ≈500 N

Number of times this force is greater than the weight =2.58x10²⁵/500

=5.16x10²²

It shows how powerful the electric force is compared to the gravitational force.   

   11. Consider a gold nucleus to be a sphere of a radius of 6.9 fermis in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at the largest separation. why do these protons not fly apart under this repulsion?

Answer: The protons at the largest separation will be at the ends of a diameter, d =2*6.9 fermi =13.8 fermi.

The force of repulsion between them will = k*q²/d²

=9x10⁹*(1.602x10⁻¹⁹)²/(13.8x10⁻¹⁵)²

=0.12x10 N

=1.2 N.               

     Though it is a very big force considering the size of protons, they do not fly apart in a nucleus because there are short-range attractive nuclear forces acting there, which overcome this repulsion.

  

   12. Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other during rubbing?   

Answer: Suppose 'n' number of electrons are transferred from one sphere to another. The electron receiving sphere will have a negative charge of,

q =n*1.602x10⁻¹⁹ C, and

the losing sphere will have a positive charge of,

q =n*1.602x10⁻¹⁹.

Separation between the spheres, r =1 cm =0.01 m.

The force between the sphere,

F =0.1 N =k*q²/r² N

→0.1 =9x10⁹*(n*1.602x10⁻¹⁹)²/0.01²

→ n²*23.10x10⁻²⁴ = 1

→n² =4.3x10²²

→n =2.07x10¹¹ electrons.             

  

   13. The NaCl molecule is bound due to the electric force between the sodium and the chloride ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75x10⁻⁸ cm, find the force of attraction between them. State the assumptions (if any) that you have made.

Answer: From the problem,

Separation, r =2.75x10⁻⁸ cm 

=2.75x10⁻¹⁰ m,

Negative charge on the chloride ion, 

q =1.602x10⁻¹⁹ C

The positive charge on the sodium ion, q =1.602x10⁻¹⁹ C

Hence the force of attraction between the ions,

F =kq²/r²

=9x10⁹*(1.602x10⁻¹⁹)²/(2.75x10⁻¹⁰)² N

=3.05x10⁻⁹ N.

Here we have assumed that the shape of both ions is spherical and the remaining electrons on both ions have no repulsive force on both the ions.            

  

   14. Find the ratio of the electric and gravitational forces between two protons.

Answer: Let the distance between them =r. Then the ratio of the electric and the gravitational force between two protons,

=F/P =(kq²/r²)/(Gm²/r²)

=kq²/Gm²

=9x10⁹*(1.602x10⁻¹⁹)²/6.67x10⁻¹¹*(1.672x10⁻²⁷)²

=1.24x10³⁶.               

  

   15. Suppose an attractive force acts between two protons, which may be written as F = Ce^-kr/r². (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi⁻¹ and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.

Answer: (a) Given that,             

F = Ce^-kr/r²

     Since the raised power to e will have a numeric value, it will not have a unit. Since r has a unit of length, k will have a unit of "per unit length" (m⁻¹). Thus the dimensional formula of k =[L⁻¹]

   Also, 'e' has a numeric value. hence the dimensional formula of C

=[dimensional formula of F]*[dimensional formula of r²]

=[MLT⁻²]*[L²]

=[ML³T⁻²]  

Its unit will be -(unit of force)*(unit of length)²

→N-m².  

(b) The repulsive force between the protons, F =k'q²/r²,

This is equal to the attractive force between them in this case, which is, 

F = Ce^-kr/r², equating we get

 Ce^-kr/r² = k'q²/r²

→C = k'q²/e^-kr

=9x10⁹*(1.602x10⁻¹⁹)²/e⁻¹*⁵

=3.4x10⁻²⁶ N-m².

 

 

 

   16. Three equal charges, 2.0x10⁻⁶ C each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two. 

Answer: The repulsive force between the charge under consideration and one of the other charges,

F = kq²/r²

=9x10⁹*(2x10⁻⁶)²/(0.05)²

=14.40 N

Diagram for Q-16

 

Since the triangle is equilateral, the angle between the two repulsive forces, each F =14.40 N, will be equal to 60°. Resultant of these two forces,

R =√(2F²+2F².cos60°)

=F√{2(1+0.5)} 

=14.40√3 N

=24.9 N              

Since the forces F are equal, this resultant will make equal angles with each force and will be along the angle bisector, i.e. the direction of the resultant will make 60°/2 =30° with extended sides from the charge under consideration.

 

  

   17. Four equal charges 2.0x10⁻⁶ C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb force experienced by one of the charges due to the remaining three.

Answer: Let four charges each of q be placed at four corners of a square ABCD.

Diagram for Q-17 

 Consider the forces on the charge at C. The repulsive forces from charges at B and D will be each equal to,

F =kq²/a²

{along the extended sides through C}

The repulsive force from the charge at A will be,

F' =kq²/(√2a)² =kq²/2a² =F/2

{along the extended diagonal through C}

      Since the equal forces, F, have a 90° angle between them, the resultant of these two will be equal to √2F, which will make 45° angle from each F, i.e. it is along the extended diagonal AC. The third force F' is also acting along this line. Hence, the resultant of all three forces,

R =√2F +F' =√2F +F/2

=1.91*F

=1.91*kq²/a²

=1.91*9x10⁹*(2x10⁻⁶)²/(0.05)² N

=27.5 N

  

    At 45° with the extended sides of the square from the charge under consideration.

  

   18. A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom (1 angstrom = 10⁻¹⁰ m and is abbreviated as Å) with the proton at the center. The hydrogen atom is said to be in the ground state in this case. Find the magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state.

Answer: The electric charges on a proton and an electron are the same in magnitude but opposite in nature. So they attract each other. And the magnitude of this force is,

F =kq²/r²

=9x10⁹*(1.6x10⁻¹⁹)²/(0.53x10⁻¹⁰)²

=82.0x10⁻⁹ N

=8.2x10⁻⁸ N.              

 

   19. Find the speed of the electron in the ground state of a hydrogen atom. The description of the ground state is given in the previous problem. 

Answer: If the speed of the electron =v, mass =m, and r = radius, then the force between them,

F =mv²/r,

But this force is the Coulomb force, which is given as,

F =kq²/r² =8.2x10⁻⁸ N

(from previous problem)

Equating we get,

v² =(r/m)*8.2x10⁻⁸

v=√{(0.53x10⁻¹⁰/9.11x10⁻³¹)*8.2x10⁻⁸}

=√(4.77x10¹²) m/s

=2.18x10⁶ m/s.           

  

   20. Ten positively charged particles are kept fixed on the X-axis at points x = 10 cm, 20 cm, 30 cm, ...., 100 cm. The first particle has a charge 1.0x10⁻⁸ C, the second 8x10⁻⁸ C, the third 27x10⁻⁸ C, and so on. The tenth particle has a charge 1000x10⁻⁸ C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin. 

Answer: Force on the 1 C charge will be,

F = k*1*(1x10⁻⁸/0.1²+8x10⁻⁸/0.2²+ ...+1000x10⁻⁸/1.0²)

=k*100*(1+2+3+ .... +10)*10⁻⁸ N

=9x10⁹*100*(10*11/2)*10⁻⁸ N

{from summation of AP}

=495*10*100 N

=4.95x10⁵ N        

   21. Two charged particles having a charge of 2.0x10⁻⁸ C each are joined by an insulating string of length 1 m, and the system is kept on a smooth horizontal table. Find the tension in the string.

Answer: The tension in the string will be equal to the Coulomb force of repulsion between the charged particles,

F =k*q²/r²

 =9x10⁹*(2.0x10⁻⁸)²/1.0² N

 =36x10⁻⁷ N

 =3.6x10⁻⁶ N. 

 

   22. Two identical balls, each having a charge of 2.00x10⁻⁷ C and a mass of 100 g, are suspended from a common point by two insulating strings each 50 cm long. The balls are held at a separation 5.0 cm apart and then released. Find (a) the electric force on one of the charged balls. (b) the components of the resultant force on it along and perpendicular to the string. (c) the tension in the string. (d) The acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.

Answer: (a) m =100 g =0.10 kg. 

The separation, r =5.0 cm =0.05 m. 

Charge on each ball q =2.00x10⁻⁷ C

The electric force on one of the charged balls,

F =kq²/r²

 =9x10⁹*(2.0x10⁻⁷)²/0.05² N

 =0.144 N. 

Diagram for Q-22

(b) The angle between the string and vertical = α.

tan α ≈ 2.5/50 =0.05

→α =2.9°

Force on the ball:-

Horizontal, F =0.144 N

Vertical weight, mg =0.10*9.81 =0.981 N.

Resultant of these two forces, R =√(0.144²+0.981²)

=0.991 N.

This resultant makes an angle ß with the horizontal.

tan ß =mg/F =0.981/0.144 =6.81

→ß =81.6°    

The angle between R and vertical =90-81.6 =8.4°.

The angle between R and the line of the string =8.4-2.9 =5.5°

Hence, the component of R along the string line =0.991*cos5.5° =0.986 N.

But this component of R will be balanced by the tension in the string T, which also acts on the ball. So, the net component of all the forces along the string = T-0.095 =0. Zero. 

For the component of the resultant perpendicular to the string, we find components of T and R perpendicular to the string.

Component of T perpendicular to the string = T.cos90° = 0

Component of R perpendicular to the string = R.sin5.5°

 =0.991*sin5.5°=0.095 N.

Hence, the component of resultant forces on the ball perpendicular to the string

=0+0.095 =0.095 N. (Away from the other ball).   

(c) The tension in the string

T =R*cos5.5° =0.991*cos5.5°

   =0.986 N.   

 

(d) Since the ball is free to move towards perpendicular to the string at that instant, we are concerned with the component of the resultant force in this direction, which we have calculated above in (b). This force =0.095 N.

Mass of the ball =0.10 kg

Hence, the acceleration =Force/mass

=0.095/0.10 m/s²

=0.95 m/s², perpendicular to the string, and going away from the other charge.   

     

   23. Two identical pith balls are charged by rubbing against each other. They are suspended from a horizontal rod through two strings of 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0x10⁻⁸ C.

Answer: Forces on the balls will be similar because the charges on each of them will be the same in amount but opposite in nature. Consider the forces on one of the balls. There are three forces in equilibrium: the weight of the ball mg, the force of attraction between the balls F, and the tension in the string T. Let us draw a diagram,

Diagram for Q-23

In triangle ABC, AB = 20 cm

AC =(5 -3)/2 = 1 cm.

Sin α =AC/AB =1/20 =0.05

And cos α =√(1-0.05²) =0.999

Since three forces are in equilibrium, from Lami's theorem

T/sin 90° =mg/sin(90+α) =F/sin(180°-α)  

→T =mg/cos α =F/sin α ----- (i)

So T =F/0.05 =20F

 =20*9x10⁹*(2.0x10⁻⁸)²/(0.03)²

 =0.08 N

So the tension in the string =0.08N =8x10⁻² N.       

Also from (i)

mg/cos α =T

→m =(T.cos α)/g

  =(0.08*0.999)/9.81 = 0.0081 kg

  =8.1 g. It is the mass of each ball.

 

 

   24. Two small spheres, each having a mass of 20 g, are suspended from a common point by two insulating strings of length 40 cm each. The spheres are identically charged, and the separation between the balls at equilibrium is found to be 4 cm. Find the charge on each sphere.

Answer: Let the charge on each ball = q.

Separation r = 4 cm =0.04 m

Hence, the force exerted on each other,

F = kq²/r² =9x10⁹*q²/0.04² 

 =5.63x10¹²q²

Forces on one of the balls will be as shown in the diagram below,

Diagram for Q-24

sin α = 2/40 =1/20 =0.05, hence,

cos α=√(1-0.05²) =0.999.       

Since the forces are in equilibrium, they will be proportional to the sine of the opposite angles.

→F/sin(180-α) =mg/sin(90+α)

→5.63x10¹²q² =0.020*9.81*sin α/cos α

→q² =0.196*(0.05/0.999)/5.63x10¹²

       =1.744x10⁻¹⁵

→q =4.17x10⁻⁸ C.

  

   25. Two identical balls, each carrying a charge q, are suspended from a common point by two strings of equal length l. Find the mass of each ball if the angle between the strings is 2θ in equilibrium. 

Answer: The situation is similar to Q-24. Here, we deduce that the angle between the string and vertical =θ.

The separation between the charges,

r =2.lsinθ 

F =kq²/r²

Diagram for Q-25

Since three forces are in equilibrium, they will be proportional to the sine of opposite angles. 

F/sin(180°-θ) = mg/sin(90°+θ)

→F =mg.sinθ/cosθ =mg.tanθ

→m =F/g.tanθ

 =kq²/(r²g.tanθ)

 =kq²/(4l²sin²θ*g*tanθ)

 =(1/4πεₒ)q²/(4l²g.sin²θ*tanθ)

 =q²cotθ/(16πεₒgl²sin²θ)

 

 

   26. A particle having a charge of 2.0x10⁻⁴ C is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. The mass of a bob is 100 g. What charge should the bob be given so that the string becomes loose? 

Answer: Assuming that the string of the pendulum will not move sideways under the repulsive electric force, the string will get loose if the weight of the bob is equal to the repulsive force. i.e. 

F = mg

→kq*2x10⁻⁴/0.10² =0.10*9.81

q =49.05/k

 =49.05/9x10⁹ C

 =5.4x10⁻⁹ C.   

  

   27. Two particles A and B having charges q and 2q respectively are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table under electrical forces. What should be the charge on C, and where should it be clamped? 

Answer: The charges q and 2q will repel each other. So the third particle must have a charge opposite in nature and should be placed somewhere on the line joining and between them. It will exert an attractive force on both of them, and the net force on each of them will be zero.

Let the third charge  = Q, and it is placed at a distance x from the charge q. So its distance from the charge 2q =d-x.

Repulsive force by q and 2q on each other = kq*2q/d²

=2kq²/d²,

The attractive force by the third charge on q and 2q will also be equal in magnitude.

kqQ/x² = k*2qQ/(d-x)²

→2x² =(d-x)²

→√2*x = d-x

→x = d/(√2+1)

{multiplying numerator and denominator by √2-1 we get}

→x =(√2-1)d.     

Now Q =?

Equating the attractive and the repulsive forces, we get,

kqQ/x² =2kq²/d²

→Q =2qx²/d²

 =2q{(√2-1)d}²/d²    

 =2(2+1-2√2)q

=(6-4√2)q.

Since this charge on C will be opposite in nature to q and 2q, we assign it a negative sign.

Thus, Q = -(6-4√2)q.    

  

   28. Two identically charged particles are fastened to the two ends of a spring of spring constant 100 N/m and natural length of 10 cm. The system rests on a smooth horizontal table. If the charge on each particle is 2.0x10⁻⁸ C, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.

Answer: The electric force exerted by the charges on each other pulls the spring. This force is

F =kq²/r²

If the extension produced =x, 

then force on the spring =K*x =100x N. Since we assume that the extension x is small in comparison to the original length, we have r ≈ 0.10 m.  So

100x =9x10⁹*(2x10⁻⁸)²/(0.10)²

→x =3.6x10⁻⁶ m.

When we assume that the extension is small compared to the original length of the spring, we can take the separation of the charges approximately equal to 0.10 m, which simplifies the calculations. Otherwise, if r =10 +x m,  then we have to solve a quadratic equation in x. 

  

   29. A particle A having a charge of 2.0x10⁻⁶ C is held fixed on a horizontal table. A second charged particle of mass 80 g stays in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is µ =0.2. Find the range within which the charge of this second particle may lie.

Answer: The charge on the second particle may be of the same nature or opposite nature to the first particle. Depending upon the nature of the charge, the electric force on the second particle will be repulsive or attractive. In both cases, the maximum magnitude of the force allowed not to disturb the equilibrium will be equal to the maximum static friction force,

F =µ*mg =0.2*0.080*9.81 N

  =0.157 N

For this magnitude of force, let the charge on the second particle = Q.

The electric force between the charges,

 kqQ/r² = F

→Q =Fr²/kq

 =0.157*(0.10)²/{9x10⁹*2.0x10⁻⁶}

 = 8.72x10⁻⁸ C.

This is the maximum magnitude of the charge, but it can be positive or negative, as stated in the beginning. So the range of the charge on the second particle 

=±8.72x10⁻⁸ C.      

  

   30. A particle A having a charge of 2.0x10⁻⁶ C and a mass of 100 g is placed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, having the same charge and mass, be placed on the incline so that it may remain in equilibrium?  

Answer: The component of gravity along the inclined plane will try to bring the particle B down, but the electric force of repulsion will try to move it up. The equilibrium position will be at a distance r along the plane where both forces will be equal. 

Diagram for Q-30

So,

kq²/r² = mg*sin 30°

→r² =2kq²/mg

  =2*9x10⁹*(2.0x10⁻⁶)²/(0.10*9.81)

 =7.34x10⁻²   

→r = 0.27 m =27 cm from the bottom i.e., from charge A.      

   31. Two particles A and B, each having a charge Q, are placed a distance d apart. Where a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force? What is the magnitude of this maximum force?

Answer: The force on the particle with charge q at C by each of the charges at A and B,

F =kQq/r², where r is the separation between Q and q. Let the distance of charge q from the midpoint of AB = x. So,

r =√{(d/2)²+x²}

Let us draw a diagram.

Diagram for Q-31

 Suppose the angle A =angle B =ß

Now the resultant of forces on the particle at C, R =2F.cos(90°-ß)

→R =2F.sinß 

  =2F.x/r

  =2kQq/r²)*x/r

  =2kQqx/{(d/2)²+x²}^{3/2    ------(i)}

  =16kQqx/(d²+4x²)^3/2 

For the resultant R to be maximum,

dR/dx =0

→16kQq[{1*(d²+4x²)^3/2-x*12x√(d²+4x²)}/(d²+4x²)³]=0

→(d²+4x²)^3/2 =12x²√(d²+4x²)

squaring both sides, 

(d²+4x²)³ =144x⁴(d²+4x²)

→(d²+4x²){144x⁴-(d²+4x²)²} =0

If d²+4x² =0, then the value of x will be imaginary. So,

144x⁴ =(d²+4x²)²,

→12x² =d²+4x²

→8x² =d²

→x² =d²/8

→x =d/(2√2)        

So the charge q will experience the maximum force when placed at a distance d/(2√2) from the midpoint of AB on the perpendicular bisector.

The magnitude of this force, 

R = 16kQqx/(d²+4x²)^3/2

 =16kQqd/{(2√2)(d²+4d²/8)^3/2

 =16kQqd/{(2√2)*(3√3)d³/(2√2)}  

 =16kQq/(3√3)d²

Let us put the value of k,

 =16(1/4πεₒ)Qq/(3√3)d²

 =3.08Qq/4πεₒd²

 

 

    32. Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB. (a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it. (b) Assuming x<<d, show that this force is proportional to x. (c) Under what conditions will the particle C execute a simple harmonic if it is released after such a small displacement? Find the time period of the oscillations if these conditions are satisfied.

Answer: (a) The electric force experienced by the particle may be calculated as in the above problem 31, part (i),

R =2kQqx/{(d/2)²+x²}^3/2

=(2/4πεₒ)Qqx/(x²+d²/4)^3/2 

=Qqx/{2πεₒ(x²+d²/4)^3/2}

(b) Since x<<d, x² will be negligible in comparison to d². So,

 R = Qqx/2πεₒ(d²/4)3/2

   =(4Qq/πεₒd³)x

    =(Constant)*x

→R ∝ x.

(c) The particle C will execute a simple harmonic motion after the release only if the nature of the charges of Q and q are opposite, i.e the forces between particles A-C and B-C are attractive in nature.

The acceleration of the particle,

a = Force/mass =R/m

The time period of the oscillation,

T =2π√(x/a)

 =2π√(mx/R)

 =2π√{mx/(4Qq/πεₒd³)x}

 =2π√(πmεₒd³/4Qq) 

 =√(π³mεₒd³/Qq).         

 

   33. Repeat the previous problem if the particle C is displaced through a distance x along the line AB. 

Answer: (a) When the particle C is displaced by a distance x towards A (assume), the separation between A and C =d/2-x and the separation between B and C =d/2+x.

The net force experienced by the particle C is, 

R =kQq/(d/2-x)²-kQq/(d/2+x)² 

 =kQq{(d/2+x)²-(d/2-x)²}/(d²/4-x²)²

 =kQq(2dx)/(d²/4-x²)² 

 =Qq(2dx/4πεₒ)/(d²/4-x²)²

 =Qqxd/{2πεₒ(d²/4-x²)²}.

(b) When x<<d, x² is negligible in comparision to d², so now,

R =Qqxd/{2πεₒ(d²/4)²}

  =(8Qq/πεₒd³)*x

  =(A constant)*x

→R ∝ x.

(c) The condition for the particle C to execute a simple harmonic motion when released after a displacement along AB is that the nature of the charge on the particle C must be the same as the charges on A and B. 

When this condition is satisfied, time period =?

Acceleration of the particle at the displacement x,

a = R/m

 =8Qqx/mπεₒd³

Time period of the oscillation,

T=2π√(x/a)

 =2π√(mπεₒd³/8Qq)

 =√(mπ³εₒd³/2Qq).

 

 

   34. The electric force experienced by a charge of 1.0x10⁻⁶ C is 1.5x10⁻³ N. Find the magnitude of the electric field at the position of the charge. 

Answer: The magnitude of the force experienced by the charge,

F =1.5x10⁻³ N

Amount of the charge,

q = 1.0x10⁻⁶ C

Hence, the magnitude of the electric field at the position of the charge,

E =F/q

 =1.5x10⁻³/1.0x10⁻⁶ N/C

 =1.5x10³ N/C. 

     

 

   35. Two particles A and B having charges of +2.00x10⁻⁶ C and of -4.00x10⁻⁶ C respectively are held fixed at a separation of 20.0 cm. Locate the point(s) on the line AB where (a) the electric field is zero (b) the electric potential is zero.

Answer: (a) Between the points A and B, the direction of fields due to the charge will be the same and hence add up, thus it will not be zero between A and B. Suppose at point C at a distance of r from A along BA, the electric field is zero. Distance AB =d =20 cm.

The electric field at point C due to charge Q at A

=kQ/r², towards BA.

The electric field at point C due to charge Q' at B,

=kQ'/(d+r)²,  

The magnitudes of these two fields will be equal,

kQ/r² =kQ'/(d+r)²

→(d+r)²/r² = Q'/Q =2

→(d/r+1)² =2

→d/r+1 =√2

→d/r =√2-1 =0.414

→r =d/0.414 =20/0.414 cm

→r =48.3 cm from A along BA.     

(b) When we take point C between A and B:-

The electric potential at a point r due to the charge Q =kQ/r.

And at distance d-r due to the charge Q' =-kQ'/(d-r). So,

kQ/r-kQ'/(d-r) =0

→(d-r)/r =Q'/Q

→d/r-1 =2

→d/r =3

→r =d/3 =20/3 cm from A along AB. 

When we take the point C at a distance r from A along BA:-

Potential due to A = kQ/r,

Potential due to B =-kQ'/(d+r).

For zero potential,

kQ/r-kQ'/(d+r) =0

→(d+r)/r =Q'/Q =2

→d/r =2-1 =1

→r =d = 20 cm from A along BA.    

 

   36. A point charge produces an electric field of magnitude 5.0 N/C at a distance of 40 cm from it. What is the magnitude of the charge? 

Answer: The electric field at a distance r from a point charge Q is given as,

E =(1/4πεₒ)Q/r²

→5.0 =9x10⁹Q/(0.40)²

→Q =5.0*0.16/9x10⁹ C

→Q =8.9x10⁻¹¹ C. 

 

 

   37. A water particle of mass 10.0 mg and having a charge of 1.50x10⁻⁶ C stays suspended in a room. What is the magnitude of the electric field in the room? What is the direction?

Answer: Since the water particle is suspended, electric force on it must be equal to the weight of the particle and acting vertically upward. 

So, electric force, F=mg

→F =mg =(1x10⁻⁵ kg)*9.8 m/s²

→F =9.8x10⁻⁵ N 

Electric field =F/q

 =9.8x10⁻⁵/1.50x10⁻⁶ N/C 

 =65.3 N/C, upward.        

 

   38. Three identical charges, each having a value 1.0x10⁻⁸ C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the center of the triangle.

Answer: The three identical charges may be either positive or negative. Accordingly the directions of the field at the center due to the individual charges maybe towards the charge or away from the charge, as shown in the diagram below.

Diagram for Q-38

   In both cases, the magnitudes of the individual fields will be the same because the distance of the center from each vertex of an equilateral triangle is the same. Also, the angle between any two individual fields is 120°. Hence the resultant field at the center by three identical charges will be zero.

    Let us now calculate the potential at the center. If each side =a, then the distance between a vertex and the center, r =(2/3)*√{a²-(a/2)²} 

→r =(2/3)(√3/2)a =a/√3

Given, a =20 cm =0.20 m

Potential at the center by single charge,

=kQ/r

=9x10⁹*1.0x10⁻⁸/(a/√3)

=90√3/0.20 

=779 V

Hence total potential =3*779 V

=2338 V ≈2.3x10³ V.        

 

   39. Positive charge Q is distributed uniformly over a circular ring of radius R. A particle having a mass m and a negative charge q, is placed on its axis at a distance x from the center. Find the force on the particle. Assuming x<<R, find the time period of oscillation of the particle if it is released from there. 

Answer: Consider a very small element of the ring having charge =dQ.

The force on the particle at P having a negative charge =q,

dF =k.dQ.q/(AP)²

Diagram for Q-39

→dF =kdQ.q/(R²+x²)

By symmetry, the resultant force on q by the whole ring having charge Q will be along PO. i.e,

F =∫dF*cos ß

But cos ß =OP/AP =x/√(R²+x²)

So, F =∫kdQ.qx/(R²+x²)^3/2  

→F={kqx/(R²+x²)^3/2}∫dQ  

→F =kQqx/(R²+x²)^3/2  

Now, assuming x<<R, x² will be negligible in comparison to R². So,

F =kQqx/R³.

Acceleration of the particle at P, 

a =F/m =kQqx/mR³ 

Hence the time-period of the oscillation 

T =2π√x/a  

  =2π√(mR³/kQq)   

{putting k =1/4πεₒ}  

We have, T=2π√(4πεₒmR³/Qq)  

→T = √(16π³εₒmR³/Qq)  

       

 

   40. A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the center of the curvature of the semicircle. 

Answer: The charge per unit length =Q/L. 

Consider a very small length dl on the rod when bent semicircularly. Charge on it,

dQ =Qdl/L. 

Electric field at O due to dQ is,

dE =k.dQ/r²

 {but L = πr, →r =L/π} 

→dE =kπ²dQ/L² =kπ²Q.dl/L³ 

Diagram for Q-40

As we see that the horizontal components of dE due to corresponding charges dQ in each quarter are neutralized and only the vertical components add up.

So, E =∫dE.cos ß

=∫kπ²Q.dl.cosß/L³ 

=(kπ²Q/L³)∫(dl.cosß)  

To evaluate ∫dl.cosß, consider the small angle dß subtended by dl at O. So, dl =r.dß =L.dß/π, Now

E =(kπ²Q/L³)(L/π)∫cosß.dß  

  =(kπQ/L²)*[sinß]*2, {limit between 0 to π/2, hence multiplied by 2 for two quadrant}

→E=(2kπQ/L²)*[1-0]

  =2πQ/4πεₒL²

  =Q/2εₒL² 

 

   41. A 10 cm long rod carries a charge of +50 µC distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both ends of the rod.   

Answer: Let AB = l be the rod, and P is the point at a distance l from each end.

Diagram for Q-41

 Clearly, ABP is an equilateral triangle. Consider a very small length dl on the rod at a distance CD = x from the midpoint D of the rod. Let the angle DPC =ß, and the small length dl makes a very small angle dß at P. So, dl =PC*dß =x.cosecß.dß, thus the magnitude of the electric field at P due to dl is, 

dE =k(Q.dl/l)/(PC)²

  =(kQ/l)*(PC*dß)/(PC)² 

 =(kQ/l)(dß/PC)

 

Now the horizontal component of this field is neutralized by the horizontal component of the field due to the dl length on the other side of point D. Thus, only the vertical component of the field will remain for each part of the rod.

    The vertical component of the field

 =dE*cosß 

 =(kQ/l).cosß.dß/PC

 =(kQ/l),cos²ß.dß/(PC*cosß)

{multiplying numerator and denominator by cosß, and PC.cosß =PD =√3l/2}

 =(2kQ/√3l²)*cosß.dß

Now the direction of the resultant field will be towards DP, and the magnitude of the field can be found out by integrating it from ß = 0 to π/6 and multiplying by 2 for the two symmetrical parts. Thus, 

E = 2∫(2kQ/√3l²)cosß.dß

 =(4kQ/√3l²)∫cosß.dß

 =(4kQ/√3l²)[sinß], limit ß=0 to π/6

 =2kQ/√3l²

Given, Q =50x10⁻⁶ C =5x10⁻⁵ C, l =0.10 m

So, E =2x9x10⁹*5x10⁻⁵/√3(0.10)² N/C

 =5.2x10⁷ N/C.  

 

    42. Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.   

Answer: Suppose the ring has a uniformly distributed charge of q per unit length. Consider a very small length dl of the ring. The small charge on dl length =q.dl,

Diagram for Q-42

 

P is a point on the axis of the ring at a distance x from the center. The distance of the point P from the dl length, 

AP = √(R²+x²)

The electric field at P due to the charge q.dl,

dE =kq.dl/AP² 

By symmetry, the component of dE perpendicular to the axis will get neutralized for the whole ring. Hence, only the components dE.cosß will add up to get the resultant electric field, and its direction will be axial.

So, E = ∫dE.cosß

 =∫kq.dl.cosß/(R²+x²)

 =kq∫dl*x/(R²+x²)^1.5

 =kqx/(R²+x²)^1.5∫dl

 =kqx*2πR/(R²+x²)^1.5

{because ∫dl =2πR=circumference of the ring}

→E =2πkqRx/(R²+x²)^1.5

For E to be maximum, dE/dx =0,

→2πkqR*d{x/(R²+x²)^1.5}/dx =0

→{1*(R²+x²)^1.5-x*1.5√(R²+x²)*2x}/(R²+x²)³ =0

→√(R²+x²){(R²+x²)-3x²} =0 

→√(R²+x²)(R²-2x²) =0

So either √(R²+x²) = 0, or (R²-2x²)=0

But the first option gives an imaginary value of x, so it must be,

R²-2x² =0

→2x² =R²

→x² =R²/2

→x =R/√2.   

  

 

    43. A wire is bent in the form of a regular hexagon, and a total charge q is distributed uniformly on it. What is the electric field at the center? You may answer this part without making any numerical calculations.  

Answer: Each side of a regular hexagon is equal in length, and the distributed charges are also equal in this case. As we have seen in Q-41, the direction of the electric field on a point on the perpendicular bisector is along the perpendicular bisector.

Diagram for Q-43

 In this case, the center of the hexagon is on the perpendicular bisector of each side. Thus, the magnitude of the electric field at the center will be the resultant of six coplanar equal electric fields having an angle of 60° between two adjacent fields. We can observe that these form three sets of equal and opposite forces that neutralize each other, thus the net field at the center will be zero.         

 

    44. A circular wire-loop of radius 'a' carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the center due to the remaining wire.   

Answer: The circumference of the wire-loop =2πa. 

Charge per unit length =Q/2πa. 

 Hence, the charge on dL length of the loop =Q.dL/2πa.

 The electric field at the center due to this charge would be, dE = kQ.dL/2πa³.

 The electric field at the center due to a uniformly charged wire-loop is zero due to the uniformly distributed radial electric fields around the center. When a small length dL of the wire is removed, its electric field dE is withdrawn, and the net electric field at the center due to the remaining loop becomes = 0 -dE 

=-dE

Thus, the magnitude of the electric field at the center due to the remaining loop is

=kQ.dL/2πa³

=(1/4πεₒ)Q.dL/2πa³

=Q.dL/8π²εₒa³.  

But its direction will be opposite to the electric field that would have been due to the charge on dL.           

 

    45. A positive charge q is placed in front of a conducting solid cube at a distance d from its center. Find the electric field at the center of the cube due to the charges appearing on its surface.   

Answer: Due to the positive charge q a field is created around the conducting cube. The free conducting electrons are attracted opposite to the direction of the field, and they start collecting near the surface facing the charge q, and this face becomes negatively charged. Thus, the rear face becomes positively charged, and an electric field is created inside the cube, the direction of which is opposite to the external field. After a time, the inner field becomes equal but opposite in direction to the external field, and the net electric field inside becomes zero. So it is clear that the strength of the electric field at the center of the cube due to the charges appearing on the surface is equal to that due to q at the center, which is

=kq/d²

=q/4πεₒd²

But its direction is towards the positive charge q.  

 

    46. A pendulum bob of mass 80 mg and carrying a charge of 2x10⁻⁸ C is at rest in a uniform, horizontal electric field of 20 kV/m. Find the tension in the thread.   

Answer: The strength of the electric field, E = 20 kV/m.

Charge on the bob, q =2x10⁻⁸ C.

The electric force on the bob, F =qE

→F =2x10⁻⁸*20x10³ N

  =4x10⁻⁴ N

Weight of the bob, W =mg

→W =80x10⁻⁶*9.8 N

  =7.84x10⁻⁴ N

The magnitude of the tension in the thread will be equal to the resultant of F and W. The angle between F and W is 90°. Hence, the magnitude of the tension =√(F²+W²) 

=√{(7.84x10⁻⁴)²+(4x10⁻⁴)² N

=8.80x10⁻⁴ N.

 

    47. A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?   

Answer: The opposite force on the particle, F =qE,

Retardation of the particle, a =F/m

→a =qE/m,

Using the equation v² =u²-2as, 

Here, v =0, hence s =u²/2a 

→s =mu²/2qE             

Before coming to momentary rest, the particle will travel a distance mu²/2qE.

 

    48. A particle of mass 1 g and charge 2.5x10⁻⁴ C is released from rest in an electric field of 1.2x10⁴ N/C. (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis? (b) How long will it take for the particle to travel a distance of 40 cm? (c) What will be the speed of the particle after traveling this distance? (d) How much is the work done by the electric force on the particle during this period?  

Answer: (a) Mass of the particle, m = 1 g =0.001 kg.  

Weight of the particle, W =mg 

→W =0.001*9.8 N

  =9.8x10⁻³ N

Charge on the particle, q =2.5x10⁻⁴ C

Strength of the field, E =1.2x10⁴ N/C

The electric force on the particle, 

F = qE

  =2.5x10⁻⁴*1.2x10⁴ N

  =3.0 N

 The ratio F/W =3.0/9.8x10⁻³ =306.12

The electric force F is more than 300 times than the weight of the particle; hence, for an approximate analysis, the force of gravity may be neglected.

(b) Acceleration of the particle, 

a =F/m =3.0/0.001 m/s²

 =3.0x10³ m/s²

Distance, s = 40 cm = 0.40 m

Initial velocity, u =0.

Time to travel 40 cm = t =?

Using the equation s =ut +½at²

→0.40 = 0+½*3x10³*t²

→t² =2*0.40/3x10³ = 2.67x10⁻⁴

→t = 1.63x10⁻² s

(c) Let the speed of the particle after 40 cm = v. Using the equation v² =u²+2as

v² =0 +2*3x10³*0.40 =2400

→v = 49.0 m/s. 

(d) The work done on the particle during this period =Force*distance

 =3.0*0.40 J

  =1.20 J. 

               

  

    49. A ball of mass 100 g and having a charge of 4.9x10⁻⁵ C is released from rest in a region where a horizontal electric field of 2.0x10⁴ N/C exists. (a) Find the resultant force acting on the ball. (b) What will be the path of the ball? (c) Where will the ball be at the end of 2 s?   

Answer: (a) Mass of the ball, m = 100 g

→m = 0.10 kg

Weight of the ball, W =mg

→W =0.10*9.8 =0.98 N

Charge on the ball, q =4.9x10⁻⁵ C

Horizontal electric field, E=2x10⁴ N/C

The electric force on the ball,

F=qE =4.9x10⁻⁵*2x10⁴ N

  =0.98 N

So, both the forces F and W acting on the particle are equal in magnitude, but the directions are at a right angle. Hence, the resultant force on the ball,

R =√(W²+F²)

 =√(2*0.98²) N

 =0.98√2 N

 =1.4 N

tanß =0.98/0.98 =1

→ß =45°

So the resultant R will make an angle of 45° with the directions of each of F and g.            

(b) Since the ball is released from rest and the net resultant force on the ball is R, it will move along a straight line in the direction of R according to Newton's second law of motion.

(c) u =0, t = 2 s, acceleration of the ball, a =F/m 

→a =1.4/0.10 =14 m/s²

From the equation, s =ut +½at² 

→s =0 +½*14*2² =28 m.

So after the end of 2 s the ball will be at 28 m from the starting point along the direction of resultant force R.       

 

 

   

    50. The bob of a simple pendulum has a mass of 40 g and a positive charge of 4.0x10⁻⁶ C. It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude 2.5x10⁴ N/C is switched on. How much time will it now take to complete 20 oscillations?   

Answer: The time period of the pendulum, 

T = 45/20 =2.25 s.  

If the length of the pendulum =l, then from the expression for the time period,  

T =2π√(l/g) 

→2.25 =2π√(l/9.8) 

→√(l/9.8)=2.25/2π   

→l = 9.8*(2.25/2π)² =12.40/π² m

Now the upward electric force on the bob, F =qE

→F =4x10⁻⁶*2.5x10⁴ N

  =0.10 N

Mass of the bob, m = 40 g =0.040 kg

Acceleration due to the electric force F,

f =F/m =0.10/0.040 =2.5 m/s², upward. 

Now the net acceleration of the bob,

a =g -f = 9.8-2.5 = 7.3 m/s²

Now the time period, T' =2π√(l/a)

= 2π√(12.40/7.3π²) s

= 2*1.30 s = 2.6 s

Now the time taken in 20 oscillations =20*2.6 s = 52 s.    

   51. A block of mass m and having a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k as shown in Figure (29-E1). A horizontal electric field E parallel to the spring is switched on. Find the amplitude of the resulting SHM of the block. 

Figure for Q-51

Answer: The electric force on the block,

F =qE 

The amplitude, x is the maximum displacement from the mean position. At the maximum displacement, the electric force and the spring force will be equal.  

kx =qE

Hence, x =qE/k.

  

   52. A block of mass m containing a net positive charge q is placed on a smooth horizontal table that terminates in a vertical wall as shown in Figure (29-E2). The distance of the block from the wall is d. A horizontal electric field E towards the right is switched on. Assuming elastic collisions (if any) find the time period of the resulting oscillatory motion. Is it a simple harmonic motion? 

Figure for Q-52

Answer: In an SHM, the force is always directed opposite to the displacement, and also the force is proportional to the displacement. In this case, when the motion starts, the direction of force as well as the direction of displacement is the same. Also, the force is constant and thus not proportional to the displacement. So it is clear that the resulting motion is not an SHM. 

  The time period of the oscillation is the time interval after which the block comes back to its original position. Let this time be T. Since the collision is elastic, the time taken by the block to travel a distance d up to the wall will be equal to the time taken to travel from the wall to the initial position. Let this time = t. So, T =2t.

Acceleration of the block, a = F/m

→a =qE/m

Initial velocity, u = 0.

The distance traveled, s =d. 

From the equation, s =ut+½at² 

d =0*t +½(qE/m)t²

→t² =2md/qE 

→t =√(2md/qE) 

So the time period, T =2t 

→T =2√(2md/qE) =√(8md/qE).               

 

   53. A uniform electric field of 10 N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm. 

Answer: The electric field, E = 10 N/C, is vertically downward. 

Taking the downward direction positive, the displacement, dr =-50 cm =-0.50 m.

Now the change in electric potential,

dV =-E.dr 

   =-10*(-0.50) volt

   =5.0 V.             

 

   54. 12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. How much is the potential difference V_B - V_A?

Answer: Since the work is being done against the electric field, the work done by the field is negative, i.e. dW =-12 J.

The change in potential energy from A to B, dU =-dW = 12 J.           

Given, q = 0.01 C,

Hence, the potential difference,

V_B -V_A =dU/q =12/0.01 V =1200 V.  

 

   55. Two equal charges, 2.0x10⁻⁷ C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process?  

Answer: Let the midway point = A and the final point =B. The magnitude of each charge, q = 2x10⁻⁷ C.  

Potential at A due to the two equal fixed charges, V' =2*kq/r 

=2kq/0.10 =20kq volts. 

The potential at B due to the two equal fixed charges, V" =2*kq/0.20 =10kq volts.

Potential difference,

 dV =V_B -V_A =10kq-20kq =-10kq,

Change in potential energy between B and A, dU = q.dV =-10kq²

Hence, the work done by the electric field, dW =-dU =10kq² 

=10*9x10⁹*(2.0x10⁻⁷)²

=90*4*10⁻⁵ J 

=3.6x10⁻³ J.         

           

 

   56. An electric field of 20 N/C exists along the X-axis in space. Calculate the potential difference V_B-V_A, where the points A and B are given by,

(a) A = (0,0); B = (4 m, 2 m)

(b) A = (4 m, 2 m); B = (6 m, 5 m)

(c) A = (0,0); B = (6 m, 5 m).

Do you find any relation between the answers of parts (a), (b), and (c)? 

Answer: The potential difference V_B-V_A is given as the negative of the dot product of E and the displacement dr (from A to B). i.e., 

dV =-E.dr, 

Given, E = i20 N/C. 

(a) dr = i4 +i2 m 

So, dV =-(i20).(i4+j2) volts 

→dV =-(20*4) volts = -80 V. 

(b) dr = i(6-4) +j(5-2) m

       =i2 + j3 m  

So, dV =-(i20).(i2 + j3)  

        =-20*2 = -40 V. 

(c) dr = i6 + j5 m 

So, dV =-(i20).(i6 + j5)

          =-20*6 =-120 V. 

        

There is a clear relation among the answers of parts (a), (b), and (c). The potential difference between two points (0,0) and (6 m, 5 m) is equal to the sum of the PDs between (0,0) -(4 m, 2 m) and (4 m, 2 m) -(6 m, 5 m). i.e. -120 V =(-80 V) + (-40 V).        

 

   57. Consider the situation of the previous problem. A charge of -2.0x10⁻⁴ C is moved from point A to point B. Find the change in electrical potential energy U_B -U_A for the cases (a), (b), and (c). 

Answer: The change in electrical potential energy dU =qdV. 

(a) Change in electrical potential energy  between A(0, 0) and B(4 m, 2 m) =qdV = (-2.0x10⁻⁴)*(-80) J

= 0.016 J.

(b) Change in electrical potential energy between A(4 m, 2 m) and B(6 m, 5 m) =qdV = (-2.0x10⁻⁴)*(-40) J 

=0.008 J. 

(c) Change in electrical potential energy between A(0, 0) and B(6 m, 5 m) =qdV =(-2.0x10⁻⁴)*(-120) J 

=0.024 J.            

 

   58. An electric field E = (i20 + j30) N/C exists in the space. If the potential at the origin is taken to be zero, find the potential at (2 m, 2 m). 

Answer: Change in position vector of the given point P(2 m, 2 m) from the origin, dr = i2 +j2 m

Potential at P, dV =-E.dr 

=-(i20 + j30).(i2 + j2) 

=-(20*2 +30*2) V  

=-100 V.               

 

   59. An electric field E = iAx exists in space A = 10 V/m². Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin. 

Answer: Here dr = i(0-10) +j(0-20) m

→dr =-i10 -j20 m =idx +jdy 

Given electric field, E =iAx

Since the electric field is a function  of the x-coordinate, totalPotential at the origin, 

dV =-E.dr 

→dV =-(iAx).(idx +jdy) =-Axdx  

Integrating,

∫dV =-∫Axdx 

→V =-A∫xdx =-A[x²/2]

The limits of integration x = 10 m to 0. 

→V = -A(0²/2 -10²/2) V

     = 100A/2 V   

{Put the value of A =10 V/m²} 

→V = 100*10/2 V =500 V.  

  

  

   60. The electric potential existing in space is V(x, y, z) = A(xy +yz +zx). (a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 SI units, find the magnitude of the electric field at (1 m, 1 m, 1 m).   

Answer: (a) V(x, y, z)=A(xy+yz+zx) 

In terms of units, 

Volts = A*m²

→A = V/m² =[ML²T⁻³I⁻¹]/[L²]

      =[MT⁻³I⁻¹]. 

(b) Since dV =-E.dr,

→E =-dV/dr 

But V and dr are functions of three variables, hence in terms of partial differentiation,

E =-i∂V/∂x -j∂V/∂y -k∂V/∂z

 =-i(y+z)A -j(z+x)A -k(x+y)A

 =-A{i(y+z) +j(z+x) +k(x+y)}

(c) Given A = 10 SI units, hence the electric field at the point (1 m, 1 m, 1 m),

E = -10{i2 +j2 +k2} N/C

The magnitude of the electric field,

=-10*√(2²+2²+2²) N/C

=-10*√12 N/C 

=-10*2√3 N/C

=-34.64 N/C

≈ -35 N/C.  

   61. Two charged particles, having equal charges of 2.0x10⁻⁵ C each, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.          

Answer: The increase in electric potential energy is equal to the negative of the work done by the electric forces. The force between two charged particles,

F = kqq'/r², here q =q', so 

F =kq²/r²

Work done for a displacement dr,

dW = kq²dr/r²

Hence, the increase in potential energy in this case, 

U = -W =∫-dW

  =-∫kq²dr/r², 

{Limits r=infinity to r=10 cm i.e. 0.10 m}  

  =-kq²[-1/r]

  = kq²[1/r], put the values and limits,

  =9x10⁹*(2x10⁻⁵)²[1/0.10-1/∞]

  =36 J.

 

 

   62. Some equipotential surfaces are shown in Figure (29-E3). What can you say about the magnitude and direction of the electric field?         

The figure for Q-62

Answer: (a) All parallel equipotential lines make an angle of 30° from the x-axis. So the line of action of the electric field will be perpendicular to these lines. Since E=-dV/dr, it is maximum in this perpendicular direction. It will be directed in the direction in which the potential decreases at the maximum rate.    

The distance between given parallel equipotential lines, dr = (10 cm)*sin 30° =5.0 cm =0.05 m. The difference of volts between consecutive equipotential surfaces, dV = 10 V. Hence, the magnitude of E =dV/dr =(10 V)/(0.05 m) =200 V/m, and direction =30° +90° =120° from the positive direction of the X-axis.

(b) Suppose the charge at the center is Q. The potential at 10 cm from it is 60 V. Thus,

kQ/x = 60

→kQ/0.10 =60

→kQ = 6 V-m  

Now consider a point P at a distance r from Q, as shown in the figure below, here

Field E =kQ/r² =6/r² V/m, where r is in meters and obviously decreases with increasing r. The direction of the field will be perpendicular to the equipotential surface, which is radially outward here.  

 

 

   63. Consider a circular ring of radius r, uniformly charged with linear charge density λ. Find the electric potential at a point on the axis at a distance x from the center of the ring. Using this expression for the potential, find the electric field at this point.         

Answer: The distance of the point from the ring circumference, R =√(r²+x²). The total charge on the ring, Q =2πrλ. Hence, the potential at the given point,

V =kQ/R 

 = (1/4πεₒ)*2πrλ/√(r²+x²) 

 = rλ/2εₒ√(r²+x²) 

The field at a point is a vector. Here, the field, E' at the point P due to a unit length of the charged ring, will be directed away along the length R. The component of E' perpendicular to the axis will be neutralized by the full ring of the charge; only the component along the axis will be added, and the resultant direction of the field will be axial and away from the ring. 

Diagram for Q-63

The magnitude of the field,

E =E'.cosß 

 =k*2πrλ*cosß/R²   

 =(1/4πεₒ)*2πrλ*(x/R)/R²  

 =rλx/2εₒR³  

 =rλx/2εₒ(r²+x²)^3/2

 

 

   64. An electric field of magnitude 1000 N/C is produced between two parallel plates having a separation of 2.0 cm as shown in Figure (29-E4). (a) What is the potential difference between the plates? (b) With what minimum speed should an electron be projected from the lower plate in the direction of the field so that it may reach the upper plate? (c) Suppose the electron is projected from the lower plate with the speed calculated in part (b). The direction of projection makes an angle of 60° with the field. Find the maximum height reached by the electron.         

The figure for Q-64

Answer:  (a) Given E =1000 N/C. dr=2cm = 0.02 m. Hence, the potential difference between the plates,  

dV =E.dr =1000*0.02 =20 V.  

(b) Since the force on an electron will be opposite to the direction of the field, it will get retarded when projected from the lower plate. Force on the electron,

F =eE  

Retardation, a =F/m =eE/m, where e and m are the charge and mass of the electron. 

Let the required initial velocity = u.  

Final velocity, v = 0. From the equation,

v² = u² -2ax, we have,    

0 = u² -2*(eE/m)*0.02 

→u² =0.04*(1.6x10⁻¹⁹*1000/9.1x10⁻³¹)

→u =√(7.03x10¹²)

→u =2.65x10⁶ m/s  

 

(c) The component of the velocity along the direction of the field =u.cos 60° =u/2 =1.32x10⁶ m/s. 

Maximum height achieved =u²/2a 

= u²m/2eE

 =(1.32x10⁶)²*9.1x10⁻³¹/(2*1.6x10⁻¹⁹*1000)

 =5.0x10⁻³ m

 =0.50 cm

 

                       

  

   65. A uniform field of 2.0 N/C exists in space in the x-direction. 

(a) Taking the potential at the origin to be zero, write an expression for the potential at a general point(x, y, z).

(b) At which points is the potential 25 V?

(c) If the potential at the origin is taken to be 100 V, what will be the expression for the potential at a general point?

(d) What will be the potential at the origin if the potential at infinity is taken to be zero? Is it practical to choose the potential at infinity to be zero?          

Answer: (a) Electric field, E =i2 N/C.

Change in position vector from origin to the point (x, y, z), dr =ix+jy+kz 

Hence, the change in potential,

dV = -E.dr  

=-(i2).(ix+jy+kz) 

=-2x V/m  

(b) The points where the potential is 25 V will be,

-2x = 25 

→x =-12.5 m,  

It is the equation of a plane that is parallel to the yz plane at a distance of -12.5 m from it.  

(c) If the potential at the origin is 100 V, then adding the difference dV to it will give the potential at a general point, i.e.,   

V+dV = (100 -2x) V/m,

As dV =-2x V/m from (a) above.                

(d) If the potential at infinity is zero, then for the origin, dr =infinity. E is constant. Hence, the change in potential

dV =-E.dr =-E*infinity =infinity.

So the potential at the origin, in this case, will be infinity, and it is not practical to choose the potential at infinity = 0 in this case.   

 

   66. How much work has to be done in assembling three charged particles at the vertices of an equilateral triangle as shown in Figure (29-E5)?          

The figure for Q-66

Answer: When the charges are far - far away, the potential energy of the system is zero. In the above configuration, the potential energy of the system is,

P =kqq'/r+kq'q"/r+kq"q/r, where,

q =2.0x10⁻⁵ C

q' =3.0x10⁻⁵ C

q" =4.0x10⁻⁵ C.

r = 0.10 m, Hence

P =(k/r)(qq'+q'q"+q"q)

 =(9x10⁹/0.10)*(2*3+3*4+4*2)*10⁻⁵*10⁻⁵

=234 J

So the mechanical energy of the system increases from zero to 234 J, which means it will require 234 J to assemble the three charges in the above configuration.

 

   67. The kinetic energy of a charged particle decreases by 10 J as it moves from a point at a potential of 100 V to a point at a potential of 200 V. Find the charge on the particle.         

Answer: Let the charge on the particle =q. The change in the kinetic energy of the particle = work done on it.

The kinetic energy reduces, which means the force on the particle is opposite to the movement. The work done by the force =(200-100)q. Equating,

(200-100)q =10

→q =10/100 =0.10 C.       

  

   68. Two identical particles, each having a charge of 2.0x10⁻⁴ C and mass of 10 g, are kept at a separation of 10 cm and then released. What would be the speeds of the particles when the separation becomes large?       

Answer: The potential energy of two particles =kq²/r

=(9x10⁹)*(2.0x10⁻⁴)²/0.10 J

=3600 J

When the separation is large, the speeds of the particles, v, will be the same because they are identical. The total kinetic energy of both particles =2*½mv² =mv². Total potential energy will be converted to kinetic energy, hence

(10/1000)*v² =3600

→v² =3600*100

→v =60*10 =600 m/s.

       

 

   69. Two particles have equal masses of 5.0 g each and opposite charges of +4.0x10⁻⁵ C and -4.0x10⁻⁵C. They are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.          

Answer: The potential energy at 1 m separation P =kq(-q)/r =-kq². 

The potential energy at 0.50 m separation P' =-kq²/0.5 =-2kq².

Reduction in P.E. =-kq²-(-2kq²) 

=kq².

This change in P.E. will be added to the K.E. of the particles. Let the speed of each particle =v. Then,

2*½mv² =kq²

→v² =kq²/m

→v =√(kq²/m)

 =√{9x10⁹*(4x10⁻⁵)²/0.005} m/s

 =√2880 m

 =53.7 m ≈54 m/s.                            

 

   70. A sample of HCl gas is placed in an electric field of 2.5x10⁴ N/C. The dipole moment of each HCl molecule is 3.4x10⁻³⁰ C-m. Find the maximum torque that can act on a molecule.        

Answer: Given, the dipole moment of an HCl molecule, p =3.4x10⁻³⁰ C-m,

Electric field, E =2.5x10⁴ N/C

Hence, the magnitude of the torque,

 =pEsinß

Since the maximum value of sinß =1, hence, the maximum torque on a molecule = pE

=3.4x10⁻³⁰*2.5x10⁴ N-m.

=8.5x10⁻²⁶ N-m   

   71. Two particles A and B, having opposite charges 2.0x10⁻⁶ C and -2.0x10⁻⁶ C, are placed at a separation of 1.0 cm. (a) Write down the electric dipole moment of this pair. (b) Calculate the electric field at a point on the axis of the dipole 1.0 m away from the center. (c) Calculate the electric field at a point on the perpendicular bisector of the dipole and 1.0 m away from the center. 

Answer: (a) Given q = 2.0x10⁻⁶ C. d = 1.0 cm =0.01 m.

Hence, the electric dipole moment of the pair, 

p =qd

 =2.0x10⁻⁶*0.01 C-m

 =2.0x10⁻⁸ C-m.

(b) The electric field at a point on the dipole axis is given as,

E =2kp/r³, where r is the distance of the point from the center of the dipole.

=2*9x10⁹*2.0x10⁻⁸/1³ 

=360 N/C. Along the axis.

(c) On the perpendicular bisector of a dipole, the electric field at a point is given as,

E =kp/r³

 =9x10⁹*2.0x10⁻⁸/1³ 

 =180 N/C. Along the antiparallel to the dipole axis.          

 

   72. Three charges are arranged on the vertices of an equilateral triangle as shown in the figure (29-E6). Find the dipole moment of the combination.  

The figure for Q-72

Answer: ABC is an equilateral triangle. At A, there is 2q charge, while at B and C, there is -q charge each. It can be assumed that there are two dipoles along AB and AC. The dipole moment of AB =qd along BA, and that of AC =qd along CA.

The diagram for Q-72

 The resultant of these two vectors is the dipole moment of the combination; its magnitude is 

p (resultant) = √{2(qd)²+2(qd)².cos 60°}

=√{3(qd)²}

=qd√3, along the angle bisector, at A and away from the triangle, as shown in the diagram.    

 

   73. Find the magnitude of the electric field at the point P in the configuration shown in the (29-E7) for d>>a. Take 2qa = p.  

The figure for Q-73

Answer: (a) The magnitude of the electric field at P =kq/d² 

   =q/4πεₒd².

(b) The two charges q and -q are equivalent to a dipole with separation r=2a. Its dipole moment, p =q*2a =2aq

P is at the perpendicular bisector of the dipole. Hence, the magnitude of the electric field here,

E =kp/d³

 =(1/4πεₒ)p/d³ 

=p/4πεₒd³.

(c) Here we take the charges q and -q separated by 2a distance apart as a dipole. Its dipole moment, p=2aq. Point P is at the perpendicular bisector of this dipole, and as in (b) above, the magnitude of the electric field at P due to this dipole, E = p/4πεₒd³. Its direction is horizontal.

The electric field at point P due to the middle charge q,

E' =q/4πεₒd². Directed vertically upward. So the electric fields E and E' are mutually perpendicular. Hence the magnitude of the resultant electric field =√(E²+E'²)

=(1/4πεₒd²)√(q²+p²/d²)   

=(1/4πεₒd³)√(q²d²+p²).         

 

 

   74. Two particles carrying charges -q and +q and having equal masses m each,  are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at one end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is slightly tilted and then released. Neglecting gravity, find the time period of small oscillations.  

Answer: Neglecting gravity, the dipole may be assumed as a pendulum and the force on the pendulum, F =qE as shown in the diagram.

Diagram for Q-74

The length of this pendulum = a. 

Mass of the bob = m, hence 

acceleration =F/m =qE/m.

The time period of a pendulum is,  

T =2π√(l/g), here, l → a, and g →qE/m

So the time period is,

T =2π√{a/(qE/m)}

  =2π√(ma/qE).                

 

   75. Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire having a mass of 6.4 g (take the atomic weight of copper to be 64 g/mol).  

Answer: Number of moles of copper in the given wire =6.4/64 =0.10 mole. 

Number of atoms in 1 mole =6.02x10²³

{Avogadro's number}

Number of atoms in the wire that have 0.10 mole =6.02x10²²

Since one atom contributes one free electron, hence 6.02x10²² atoms will contribute 6.02x10²² free electrons. 

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CHAPTER- 13 - Fluid Mechanics 


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CHAPTER- 36 - Permanent Magnet

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CHAPTER- 38 - Electromagnetic Induction

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EXERCISES - Q1 -Q10

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EXERCISES - Q61 - Q70

EXERCISES - Q71 - Q80

EXERCISES - Q81 - Q90

EXERCISES - Q91 - Q98

 

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EXERCISES - Q1 - Q10

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CHAPTER- 42 - Photoelectric Effect and Wave-Particle Duality

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CHAPTER- 43 - BOHR'S MODEL AND PHYSICS OF THE ATOM

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EXERCISES - Q1 - Q10

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CHAPTER- 44 - X-rays

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EXERCISES - Q21 - Q27

CHAPTER- 45 - Semiconductors and Semiconductor Devices

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EXERCISES - Q1 - Q10

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CHAPTER- 46 - The Nucleus

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 - Q40

EXERCISES - Q41 - Q53

CHAPTER 47 - The Special Theory of Relativity

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES -Q11 - Q28

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