Wednesday, October 28, 2020

H C Verma solutions, ELECTRIC FIELD AND POTENTIAL, EXERCISES, Q51 to Q60, Chapter-29, Concepts of Physics, Part-II

Electric Field and Potential


EXERCISES, Q51 to Q60


   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 which 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 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, 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 VB - VA?



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,

VB -VA =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 =VB -VA =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 VB-VA 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 VB-VA is given as negative of the dot product of E and 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 UB -UA 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 x- cordinate, 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 variable 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

  

------------------------------------------------

Click here for all links → kktutor.blogspot.com 

===<<<O>>>===


My Channel on YouTube  →  SimplePhysics with KK


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

Click here for → Questions for Short Answer 

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)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

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".

---------------------------------------------------------------------------------

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)

-------------------------------------------------------------------------------

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


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


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"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

Posted by at
Labels: , , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)