Sunday, February 28, 2021

H C Verma solutions, Capacitors, EXERCISES, Q31-Q40 Chapter-31, Concepts of Physics, Part-II

Capacitors

EXERCISES, Q31 - Q40


   31. A charge of 20 µC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10 µF. Calculate the potential difference developed between the plates. 



Answer: Since the capacitors are isolated, the charge of 20 µC given to one of the plates will get distributed to each of the faces of this plate. So inner face of this plate has 10 µC charge. The charge on the other plate is zero, but the 10 µC charge on the first plate induces -10 µC charge on the inner surface of the other plate. To keep the total charge zero on this plate the outer surface gets charge of 10 µC. So the inner surfaces of the plates have 10 µC and -10 µC of charge. Thus the charge on the effective capacitor is q =10 µC.

      If we understand the above process, the charge on the capacitor can be directly calculated as q =(20-0)/2 =10 µC. 

  Given capacitance C = 10 µF. 

The potential difference developed between the plates, V = q/C 

→V =(10 µC)/(10 µF) = 1 volt.       





    32. A charge of 1 µC is given to one plate of a parallel plate capacitor of capacitance 0.1 µF and a charge of 2 µC is given to the other plate. Find the potential difference developed between the plates.  



Answer: Given that charge on one plate q'=1 µC and the charge on the other plate q" =2 µC. 

As discussed in the previous problem, the effective charge on the capacitor can be directly calculated as,

q =(q"-q')/2 =(2-1)/2 µC =0.5 µC.  

Hence the potential difference developed between the plates,

V =q/C =(0.5 µV)/(0.1 µF) =5 volts       





 

   33. Each of the plates shown in figure (31-E19) has a surface area (96/εₒ)x10⁻¹² F-m on one side and the separation between the consecutive plates is 4.0 mm. The emf of the battery connected to is 10 volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it. 
The figure for Q-33



Answer: Area of each surface, A =(96/εₒ)x10⁻¹² F-m. Separation between the plates, d =4.0 mm =0.004 m. Four plates in the figure make three capacitors. The capacitance of each of the capacitor, C =εₒA/d

→C =εₒ*(96/εₒ)x10⁻¹²/0.004 F 

→C =24x10⁻⁻⁹ F =2.4x10⁻⁸ F

Since each plate is not connected to the battery terminal, it is not a parallel combination. The second and fourth plates are isolated, hence it is a series combination. So the equivalent capacitance C' is given as

1/C' =1/C +1/C +1/C = 3/C

→C' =C/3 =2.4x10⁻⁸/3 F =0.8x10⁻⁸ F

→C' =8x10⁻⁹ F

Potential difference, V =10 volts

Hence the charge on the equivalent capacitor, 

Q = CV =8x10⁻⁹*10 =8x10⁻⁸ C 

Diagram for Q-33


In the series connection, the charge on the equivalent capacitor =the charge on the individual capacitor. 

Hence the charge on each capacitor, also Q =8x10⁻⁸ C 

From the figure the charge on each connected plate (1st and 3rd plate) =2Q =2*8x10⁻⁸ C =0.16x10⁻⁶ C =0.16 µC.    

     





 

   34. The capacitance between the adjacent plates shown in figure (31-E20) is 50 nF. A charge of 1.0 µC is placed on the middle plate. (a) What will be the charge on the outer surface of the upper plate? (b) Find the potential difference developed between the upper and the middle plates. 
The figure for Q-34



Answer: (a) The capacitance between upper and middle plate, C =50 nF.

Charge on the upper plate = zero,

Charge on the middle plate =1.0 µC,

Hence the charge on the capacitor comprising the upper two plates,

Q =(1 µC -0)/2 =0.50 µC,

It means, the charge on the upper surface of the middle plate =0.50 µC and the charge on the lower surface of the upper plate = -0.50 µC,

Since the total charge on the upper plate is zero, the charge on the outer surface of the upper plate must be opposite of -0.50 µC, ie. = 0.50 µC,


(b) The potential difference developed between the upper and middle plates,

V =Q/C =(0.50 µC)/(50 nF)

   =(0.50x10⁻⁶ C)/(50x10⁻⁹ F)

   =0.01x10³ V

   =10 V.           







 

   35. Consider the situation of the previous problem. If 1.0 µC is placed on the upper plate instead of the middle, what will be the potential difference between (a) the upper and the middle plates and (b) the middle and the lower plates? 



Answer: (a) Same as in the solution of the previous problem, the charge on the capacitor of the upper two plates,  

Q =(1-0)/2 =0.50 µC

Hence the potential difference developed between the upper two plates,

V =Q/C =(0.50 µC)/(50 nF) =10 V.


Diagram for Q-35

(b) The charge distribution on plate surfaces is shown in the diagram above. Opposite surfaces will develop opposite charges. Since the middle and the lower plates have total charges equal to zero, both sides of these plates will also have equal and opposite charges.

     As we see the charge on the capacitor of the lower two plates is also equal to Q =0.50 µC. Hence the potential difference developed between middle and lower plates, V =Q/C 

→V =(0.50 µC)/(50 nF) =10 V.             







 

   36. Two capacitors of capacitances 20.0 pF and 50 pF are connected in series with a 6.00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.  



Answer: Given C' =20.0 pF, C" =50 pF

V =6.00 V,

Equivalent capacitance in series, C is given as

1/C =1/C' +1/C"

→1/C =1/20 +1/50 = (5+2)/100

→C =100/7 pF.

The charge on the equivalent capacitor,

Q =CV =(100/7)*6 =600/7 pC.  


(a) Since the charge on each capacitor in series combination is the same as on the equivalent capacitor, hence the charge on each of the capacitors is 

Q =600/7 pC.

So, the potential difference across 20 pF capacitor, V' =Q/C' 

            =(600/7 pC)/(20 pF)

            =30/7 Volt

            =4.29 V,


The potential difference across 50 pF capacitor, V" =Q/C"

                 =(600/7 pC)/(50 pF)

                 =1.71 V 


(b) The energy stored in a capacitor is given as, 

E =½QV

So, energy stored in 20 pF capacitor,

E' =½*(600/7 pC)*4.29 pJ

    =184 pJ

And the energy stored in 50 pF capacitor,

E" =½(600/7 pC)*1.71 pJ

    =73.3 pJ                                  



 



 

   37. Two capacitors of capacitances 4.0 µF and 6.0 µF are connected in series with a battery of 20 V. Find the energy supplied by the battery. 



Answer: Given, C' =4.0 µF, C" =6.0 µF, V =20 volts.

C' and C" are connected in series. Equivalent capacitance C is given as,

1/C =1/C' +1/C",

→C =C'C"/(C'+C") 

      =4*6/(4+6) 

      =24/10 µF

      =2.4 µF

Hence the charge on the equivalent capacitor,

Q =CV =2.4*20 µC =48 µC

In supplying this charge the battery has to do work. It moves the charge Q through 20 V. Hence the energy supplied by the battery is equal to the work done which is  =QV 

  =(48 µC)*(20 volts)

  =960 µJ.  

{Please note that the energy supplied by the battery (QV) is double the energy stored on the capacitor (½QV). You may wonder where the half energy goes. The answer is that half the energy supplied by the battery is used to heat the wires or other appliances} 

 

         



 

   38. Each capacitor in figure (31-E21) has a capacitance of 10 µF. The emf of the battery is 100 V. Find the energy stored in each of the four capacitors. 

The figure for Q-38



Answer: Capacitors b and c are connected in parallel, so equivalent capacitance of b and c is given as, 

C* =10 +10 =20 µF. 

C* is connected in series with two other capacitors each of 10 µF. Hence the equivalent capacitance of the whole arrangement C is given as 

1/C =1/10 +1/20 +1/10   

→1/C =5/20 =1/4

→C =4 µF

Potential difference =100 V, 

Hence the charge on the equivaent capacitor Q =CV =4*100 =400 µC.

This charge Q will be on each of the series-connected capacitors, i.e. 10 µF, C* and 10 µF. 

Energy stored on each of 10 µF capacitors i.e. a and d =Q²/2C =(400)²/(2*10) µJ

=8000 µJ =8 mJ.


Now the charge Q =400 µC on the capacitor C* will be equally divided to each of the 10 µF capacitors connected in parallel, i.e. Q' =200 µC. Hence the energy stored in capacitors b and c =Q'²/2C

=(200)²/(2*10) µJ

=2000 µJ

=2 mJ.                          






 

   39. A capacitor with stored energy 4.0 J is connected with an identical capacitor with no electric field in between. Find the total energy stored in the two capacitors. 



Answer: Stored energy of the first capacitor U =Q²/2C =4.0 J.

 The capacitance of both the capacitors is the same hence the charge is equally distributed between them. Charge on each capacitor =Q/2. 

    So the energy stored on each of them =(Q/2)²/2C

=Q²/8C

The total energy stored in the two capacitors =2*Q²/8C

=Q²/4C

=½*Q²/2C

=½*4 J

=2 J.            





 

   40. A capacitor of capacitance 2.0 µF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4.0 µF as shown in figure (31-E22). Find (a) the charge on each of the two capacitors after the connection. (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other. 
The figure for Q-40



Answer: (a) Charge on the 2µF capacitor before connection,

Q =(2 µF)*(12 V) = 24 µC

When it is connected to the 4 µF capacitors as in the given figure, the potential difference across the plates of each capacitor is the same, say =V'.

If Q' and Q" are charges on 2 µF and 4 µF capacitors respectively then,

V' =Q'/2, also V' =Q"/4

So, Q'/2 =Q"/4 

→Q" =2Q'

 But Q' +Q" = Q

→Q' +2Q' =24 µC

→3Q' =24 µC

→Q' = 8 µC

So the charge on 2 µF capacitor is 8 µC and on 4 µF capacitor =2*8 =16 µC.


(b) Electrostatic energy stored in 2 µF capacitor =Q'²/(2*2) =8²/4 =16 µJ.

{Since U =Q²/2C}

Energy stored in 4 µF capacitor 

=Q"²/(2*4) =16²/8 µJ

=32 µJ 


(c) Energy stored in 2 µF capacitor before connection =Q²/2C

=24²/(2*2)

=144 µJ

Energy stored in two capacitors after connection =16 µJ +32 µJ =48 µJ

The heat produced during the charge transfer =144 µJ -48 µJ

     =96 µJ.                   

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

Click here for all links → kktutor.blogspot.com 

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


My Channel on YouTube  →  SimplePhysics with KK


Links to the Chapters











EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

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 No comments:
Labels: , , ,
Subscribe to: Posts (Atom)