Monday, October 19, 2020

H C Verma solutions, ELECTRIC FIELD AND POTENTIAL, EXERCISES, Q31 to Q40, Chapter-29, Concepts of Physics, Part-II

Electric Field and Potential


EXERCISES, Q31 to Q40


   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 place 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 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 semicircular. 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 
 


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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- 5 - Newton's Laws of Motion


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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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