Sunday, October 11, 2020

H C Verma solutions, ELECTRIC FIELD AND POTENTIAL, EXERCISES, Q21 to Q30, Chapter-29, Concepts of Physics, Part-II

Electric Field and Potential


EXERCISES, Q21 to Q30


   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 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 point 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 theoram

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. Which 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 string 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 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,

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.

Electric force between the charge,

 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 the 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.

 

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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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

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Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

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Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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

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CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

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

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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