Sunday, October 25, 2020

H C Verma solutions, ELECTRIC FIELD AND POTENTIAL, EXERCISES, Q41 to Q50, Chapter-29, Concepts of Physics, Part-II

Electric Field and Potential


EXERCISES, Q41 to Q50


   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 the ends of the rod.   



Answer: Let AB = l is 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 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 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 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 net electric field inside becomes zero. So it is clear that the strength of the elctric 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 the 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 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. 

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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- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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