Friday, February 5, 2021

H C Verma solutions, Capacitors, EXERCISES, Q11-Q20 Chapter-31, Concepts of Physics, Part-II

Capacitors


EXERCISES, Q11 - Q20


   11. The outer cylinders of two cylindrical capacitors of capacitance 2.2 µF each, are kept in contact and the inner cylinders are connected through a wire. A battery of emf 10 V is connected as shown in figure (31-E5). Find the total charge supplied by the battery to the inner cylinders.
The figure for Q-11



Answer: The capacitors are connected in parallel and the inner cylinders are connected to the positive terminal of the battery. The potential difference across each capacitor is the same i.e. 10 V. Hence the equivalent capacitance of the connected capacitors,

C = 2.2 +2.2 =4.4 µF. 

Now the total charge supplied by the battery to the inner cylinder,  

Q = CV

   =(4.4 µF)x(10 V) 

   =44 µC.        






 

    12. Two conducting spheres of radii R₁ and R₂ are kept widely separated from each other. What are their individual capacitances? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series-parallel connections. 



Answer: The individual capacitance of an isolated sphere =4πεₒR.

Hence the capacitances of the given spheres are, 4πεₒR₁ and 4πεₒR₂.

         When the spheres are connected by a wire, both will come to the same potential due to the flow of charge.

                An isolated spherical capacitor is considered as a spherical capacitor having two concentric spheres in which the outer sphere is at a distance of infinity. In this case, the inner spheres (isolated spheres) are connected and at the same potential. Hence the connection is essentially a parallel combination. Thus the equivalent capacitance,

C =C₁ +C₂

   =4πεₒR₁ +4πεₒR₂

   =4πεₒ(R₁+R₂).    





 

   13. Each of the capacitors shown in figure (31-E6) has a capacitance of 2 µF. Find the equivalent capacitance of the assembly between points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors. 

The figure for Q-13


Answer: It is clear from the figure that the three capacitors in each row are connected in series and then each row is connected in parallel. If the equivalent capacitance of each row =C, then 

1/C = 1/2 +1/2 +1/2 =3/2

→C =2/3 µF.

If we replace the series combinations with the equivalent capacitances, the connection will look like below.

Diagram for Q-13

Now, these three combinations are in parallel. Hence the equivalent capacitance of the whole assembly between points A and B,  

Cₑᵩ =C +C +C =3C 

     =3*(2/3) µF

     =2 µF.  


The potential difference across each row =60 V. The equivalent capacitance of each row is C =2/3 µF. Hence the charge on the equivalent capacitor,

Q = CV =(2/3 µF)*(60 V)

           =40 µC.

The charge on each capacitor in a series combination is the same and equal to the charge on the equivalent capacitor. Hence the charge on each capacitor = 40 µC. The capacitance of each capacitor c = 2 µF. Hence the potential difference appearing on the individual capacitor, 

v =Q/c =(40 µC)/(2 µF)

   =20 volts.           





 

   14. It is required to construct a 10 µF capacitor which can be connected across a 200 V battery. Capacitors of capacitance 10 µF are available but they can withstand only 50 V. Design a combination that can yield the desired result. 



Answer: As we have seen in the previous problem that we can assemble many capacitors of capacitance C each to have an equivalent capacitance of C but with difference potential difference. In this case also we will arrange them in series and then each series in parallel to achieve our goal.

    Since one capacitor can withstand only 50 V, to connect across 200 V battery we need 200/50 = 4 capacitors in a series. If the equivalent capacitance of one such series is C', then 

1/C' = 1/C +1/C +1/C + 1/C =4/C

→C' = C/4

Suppose we need n such series to connect in parallel. Then the equivalent capacitance of the assembly,

C" =nC' 

→n =C"/C'.

But we have to keep C" = C. Hence

n = C/(C/4) = 4.

So to achieve the requirement, we need to assemble four 10 µF capacitor in series and then four such series in parallel.              






   15. Take the potential of the point B in figure (31-E7) is to be zero. (a) Find the potentials at points C and D. (b) If a capacitor is connected between C and D, what charge will appear on this capacitor?
The figure for Q-15



Answer: Equivalent capacitance C' of the upper row which is in series, is 

1/C' =1/4 +1/8 = 3/8

C' = 8/3 µF.

Potential difference across C' =50 V. 

Charge on C', Q =C'V =(8/3 µF)*(50 V)

→Q = 400/3 µC. 

This Q will appear on each of the 4 µF and 8 µF capacitors. Potential difference across 8 µF capacitor =(400/3)/(8) V

=50/3 V.

Since the potential of the point B is zero, hence the potential at C =0 +50/3

 =50/3 V.

             

Eqivalent capacitance C" of lower row which is also in series,

1/C" =1/3 +1/6 =3/6 =1/2

→C" = 2 µF.

Charge on C'', Q =C''V =(2 µF)*(50 V)

→Q = 100 µC. 

This Q will appear on each of the 3 µF and 6 µF capacitors. Potential difference across 6 µF capacitor =(100)/(6) V

=50/3 V.

Since potential at B is zero, potential at D = 0 +(50/3) = 50/3 V.

So the potential at each of the points C and D = 50/3 V.      


       Now the potential difference across C and D =50/3 -50/3 =0 V.

   Hence the charge on a capacitor connected across C and D = CV =C*0 =zero.  





 

 

   16. Find the equivalent capacitance of the system shown in figure (31-E8) between points a and b. 



Answer: Capacitors C₁ and C₂ are connected in series. If the equivalent capacitance of C₁ and C₂ is C', then

1/C' =1/C₁ +1/C₂

→1/C' =(C₁ +C₂)/C₁C₂

→C' = C₁C₂/(C₁+C₂)

Now the given arrangement can be replaced with three capacitors C', C₃ and C' connected in parallel between the points a and b. Equivalent capacitance of the system now is 

C" =C' +C₃ +C'

  =C₃ +2C'

  =C₃ +2C₁C₂/(C₁+C₂).

  





 

   17. A capacitor is made of a flat plate area A and a second plate having a stair-like structure as shown in figure (31-E9). The width of each stair is a and the height is b. Find the capacitance of the assembly. 
The figure for Q-17




Answer: The given arrangement can be assumed as three capacitors connected in parallel as shown below.    

Diagram for Q - 17

The area of each capacitor plate, 

a =A/3.

Distance between plates, 

1st capacitor - d 

2nd capacitor -d+b

3rd capacitor -d+2b

Capacitance of 1st capacitor,

C₁ =εₒ(A/3)/d =εₒA/3d

Capacitance of second capacitor,

C₂ =εₒ(A/3)/(d+b) =εₒA/{3(d+b)}

Capacitance of third capacitor,

C₃ =εₒ(A/3)/(d+2b) =εₒA/{3(d+2b)}

Eqivalent capacitance of these three capacitors in parallel,

Cₑᵩ =C₁+C₂+C₃

=εₒA/3d+εₒA/{3(d+b)}+εₒA/{3(d+2b)}

=(εₒA/3){1/d +1/(d+b) +1/(d+2b)}

{Taking LCM and adding}

=(εₒA/3){(d+b)(d+2b)+d(d+2b)+d(d+b)}/{d(d+b)(d+2b)}

=(εₒA/3)(d²+2b²+3bd+d²+2bd+d²+bd)/{d(d+b)(d+2b)}

=(εₒA/3)(3d²+6bd+3d²)/{d(d+b)(d+2b)}

=εₒA(3d²+6bd+3d²)/{3d(d+b)(d+2b)}

   




 

   18. A cylindrical capacitor is constructed using two coaxial cylinders of the length 10 cm and of radii, 2 mm and 4 mm. (a) Calculate the capacitance. (b) Another capacitor of the same length is constructed with cylinders of radii 4 mm and 8 mm. Calculate the capacitance.  



Answer: The capacitance of a cylindrical capacitor =2πεₒl/ln(R₂/R₁)

where l = length of the cylinder,

R₂ = radius of outer cylinder

R₁ = radius of inner cylinder.


(a) Here R₂ = 4 mm, R₁ =2 mm, 

l =10 cm =0.10 m

Hence ln(R₂/R₁) = ln(2)

So the capacitance here,

C =2πεₒl/ln(2)

 =2π*8.85x10⁻¹²*0.10/0.693 F

 =8.0x10⁻¹² F

 =8.0 pF



(b) Now, R₂ =8 mm, R₁ =4 mm, 

l =10 cm =0.10 m

Hence ln(R₂/R₁) =ln(2) =0.693

So the capacitance in this case,

C =2πεₒl/ln(R₂/R₁)

  =2π*8.85x10⁻¹²*0.10/0.693

  =8x10⁻¹² F

  =8.0 pF. 





 

   19. A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor?



Answer: C =100 pF =100x10⁻¹² F

Charge on the capacitor, Q =CV

→Q =100x10⁻¹²*24 C

      =2.4x10⁻⁹ C.

When this is now connected to a 20 pF capacitor, some of the charges go to this small capacitor but now both capacitors are at the same potential say at V'.

     Suppose the 20 pF capacitor gets charge = Q'. So the charge on the 100 pF capacitor = Q-Q'.

So for the first capacitor 

V' =(Q-Q')/(100x10⁻¹²)

and for the second capacitor,   

V' =Q'/(20x10⁻¹²),

Equating we get,

(Q-Q')/5 =Q'/1

→5Q' =Q -Q' 

→6Q' =Q

→Q' =Q/6 =2.4x10⁻⁹/6

     =4x10¹⁰ C

Hence the common new potential difference V' =4x10¹⁰/20x10⁻¹² volts.

→V' =20 V.               





 

   20. Each capacitor shown in figure (31-E10) has a capacitance of 5.0 µF. The emf of the battery is 50 V. How much charge will flow through AB if the switch S is closed? 
The figure for Q-20



Answer: When S is open we have two capacitors in parallel equivalent to 2C and this is connected in series with the capacitor C near switch. So the equivalent capacitance of the assembly C' is given by

1/C' =1/2C +1/C =3/2C

→C' =2C/3

Hence the charge given by the battery,

Q =(2C/3)*V

   =(2*5/3)*50 µC

   =500/3 µC 


        Now S is closed, capacitance of the shorted capacitor becomes zero as potential difference across it is zero. Only parallel capacitors are effective. Now the equivalent capacitance =2C. Charge on it,

Q' =(2C)*V

  =2*5*50 µC 

  =500 µC.

It is more than the initial charge. Hence the extra charge will flow through AB after S is closed is,

Q' -Q =500 -500/3 µC

       =1000/3 µC

      =3.33x10² µC

      =3.33x10²x10⁻⁶ C

      =3.33x10⁻⁴ C.      

====================

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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

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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

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