Saturday, October 31, 2020

H C Verma solutions, ELECTRIC FIELD AND POTENTIAL, EXERCISES, Q61 to Q70, Chapter-29, Concepts of Physics, Part-II

Electric Field and Potential


EXERCISES, Q61 to Q70


   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 of and direction of the electric field?         
The figure for Q-62



Answer: (a) All parallel equipotential lines are making 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 and 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 consicutive 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 X-axis.


(b) Suppose the charge at the center is Q. 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 meter and obviously decreasing 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 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 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, the potential is 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 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 potential 100 V to a point at potential 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 that 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 K.E. of te particles. Let 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    

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Links to the Chapters





Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 7 - Circular Motion

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CHAPTER- 6 - Friction

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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