Wednesday, March 24, 2021

H C Verma solutions, Capacitors, EXERCISES, Q61-Q68 Chapter-31, Concepts of Physics, Part-II

Capacitors

EXERCISES, Q61 - Q68


   61. A spherical capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric constant K up to a radius c as shown in figure (31-E27). Calculate the capacitance.  
The figure for Q-61



Answer: We can think of it as if two spherical capacitors connected in series. The capacitance of the inner capacitor,

C' =4πεₒcaK/(c-a)

The capacitance of the outer capacitor,

C" =4πεₒbc/(b-c)

Equivalent capacitance C is given by

C =C"C'/(C'+C")

=[(4πεₒ)²abc²K/{(c-a)(b-c)}]/4πεₒc{Ka/(c-a)+b/(b-c)}

=4πεₒKabc/{Ka(b-c)+b(c-a)}.

 




 

   62. Consider an assembly of three conducting concentric spherical shells of radii a, b and c as shown in figure (31-E28). Find the capacitance of the assembly between points A and B.  
The figure for Q-62



Answer: It is two spherical capacitors connected in series. The capacitance of the inner capacitor, 

C' =4πεₒab/(b-a)

The capacitance of the outer capacitor,

C" =4πεₒbc/(c-b)

The equivalent capacitance,

C =C'C"/(C'+C")

=[(4πεₒ)²ab²c/{(b-a)(c-b)}]/[4πεₒb{a/(b-a)+c/(c-b)]

=4πεₒabc/{a(c-b)+c(b-a)}

=4πεₒabc/(ac-ab+bc-ac)

=4πεₒabc/{b(c-a)}

=4πεₒac/(c-a).

 




 

    63. Suppose the space between the two inner shells of the previous problem is filled with a dielectric of dielectric constant K. Find the capacitance of the system between A and B. 



Answer: It is similar to the problem 61. The capacitance of the inner spherical capacitor,

C' =4πεₒKab/(b-a)

The capacitance of the outer spherical capacitor,

C" =4πεₒbc/(c-b)

The equivalent capacitance between A and B is,

C =C'C"/(C'+C")

=[(4πεₒ)²Kab²c/{(b-a)(c-b)}]/[4πεₒb{Ka/(b-a)+c/(c-b)}]

=4πεₒKabc/{Ka(c-b)+c(b-a)}.

 




 

    64. An air-filled parallel-plate capacitor is to be constructed which can store 12 µC of charge when operated at 1200 V. What can be the minimum plate area of the capacitor? The dielectric strength of air is 3x10⁶ V/m.  



Answer: Given, Q =12 µC, V =1200 volts, hence capacitance C  =Q/V

= 12/1200 µF

= 0.01 µF

Given also the dielectric strength of air =3x10⁶ V/m. It means it is the minimum field at which the breakdown of dielectric occurs. Hence for the minimum plate area (A), we take the electric field between the plates, E = dielectric strength of the air

=3x10⁶ V/m

For a uniform field, E = V/d

→d =V/E

    =1200/3x10⁶ m

    =4x10⁻⁴ m   

Now capacitance C = εₒA/d

→A =Cd/εₒ

     =0.01x10⁻⁶*4x10⁻⁴/8.85x10⁻¹² m²

     =0.45 m². 




 

    65. A parallel plate capacitor with a plate area of 100 cm² and separation between the plates 1.0 cm is connected across a battery of 24 volts. Find the force of attraction between the plates.  



Answer: A = 100 cm² =0.01 m²

d = 1.0 cm = 0.01 m

V = 24 volts

The electric field between the plates

E = 𝜎/2εₒ =Q/2Aεₒ

The force of attraction between the plates,

F =QE =Q²/2Aεₒ

   =(CV)²/2Aεₒ

   =(εₒAV/d)²/2Aεₒ

   =εₒAV²/2d²      

 =8.85x10⁻¹²*0.01*24²/{2*(0.01)²} N
   =2.54x10⁻⁷ N.





 

    66. Consider the situation shown in figure (31-E29). The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf Ԑ. All surfaces are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.   
The figure for Q-66



Answer: Let the length of plates with dielectric = x. Area of plates =bx.

The capacitance of the portion with dielectric, C' =Kεₒbx/d

The capacitance of the portion without dielectric, C" =εₒb(l-x)/d

Since area of plates in this capacitor =b(l-x)

These two types of capacitors are connected in parallel , hence the equivalent capacitance C = C'+C"

→C =(εₒb/d){Kx+(l-x)}

→C =εₒb{l+x(K-1)}/d ----- (i)

Suppose the resultant of electric forces F pulls the dielectric inside and we consider a very very small displacenment dx along F (as shown in figure below). Due to this there will be a very small change in the capacitance, say dC. 
Diagram for Q-66

Differentiating (i) both sides with respect to x,

dC/dx =εₒb(K-1)/d --------- (ii)

Let dQ be the small charge supplied by the battery due to a change of capacitance by dC. Then,

dQ =ԐdC

The work done by the battery = ԐdQ =Ԑ²dC

Work done by the resultant of electric forces force F = -Fdx

{Negative sign is for the work done by the capacitor}

Total work-done =Ԑ²dC -Fdx

This work done should be equal to the change in energy of the capacitor. The change in energy,

dU =½(dC)Ԑ²

Equating,  

Ԑ²dC -Fdx =½(dC)Ԑ²

→Fdx =½Ԑ².dC

→F =½Ԑ²(dC/dX)

   =½Ԑ²εₒb(K-1)/d 

This force by the capacitor on the dielectric is balanced by the mass M. Hence 

Mg =F

→M =½Ԑ²εₒb(K-1)/gd.

    




 

    67. Figure (31-E30) shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths l₁ and l₂. The left half of the dielectric slab has a dielectric constant K₁ and the right half K₂. Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium. 
The figure for Q-67



Answer: Suppose the emf of the left battery = Ԑ and the emf of the right battery = Ԑ'.

     As we have derived the expression for force on the dielectric by the capacitor in the previous problem, accordingly the force on the dielectric by the left capacitor, 

F =½Ԑ²εₒb(K₁-1)/d 

The force on the dielectric by the right capacitor

F' =½Ԑ'²εₒb(K₂-1)/d

To the dielectric slab be in equilibrium,

F = F'

→½Ԑ²εₒb(K₁-1)/d =½Ԑ'²εₒb(K₂-1)/d

→Ԑ²(K₁-1) =Ԑ'²(K₂-1)

→Ԑ²/Ԑ'² =(K₂-1)/(K₁-1)

→Ԑ/Ԑ' = √{(K₂-1)/(K₁-1)}.

 

 



 


    68. Consider the situation shown in figure (31-E31). The plates of the capacitor have plate area A and are clamped in the laboratory. The dielectric slab is released from rest with a length 'a' inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.   
The figure for Q-68



Answer: The lengths 'l' of the plates and the dielectric are the same. Area of plate = A. Hence the width of the plates, b =A/l.

The force on the dielectric

F =½Ԑ²εₒb(K-1)/d 

→F =½Ԑ²εₒA(K-1)/ld

Assuming mass of the dielectric = m, its acceleration f is given as,

f = F/m

→f = ½Ԑ²εₒA(K-1)/lmd

   The force F will act on the dielectric slab till it is fully inside the capacitor i.e. upto the distance (l-a). At this instant it will have some velocity and it will partially come out at the other end. At the instant the dielectric comes slightly out of the capacitor the direction of the force gets reversed with the same magnitue (Since the magnitude of the reversed force is same, the magnitude of the retardation is also the same). This will stop the dielectric after it has travelled a distance (l-a) after fully covering the capacitor. The dielectric will again move into the capacitor with same acceleration, and as it just moves inside the capacitor after covereing it, the force and the acceleration will again get reversed. The slab will stop at its original position and the same movement will get repeated. Thus the slab will move to and fro with a periodic motion. (See diagram below ↓).
Diagram for Q-68

    Note that it is a periodic motion i.e. a motion that gets repeated after a fixed time interval but it is not a simple harmonic motion because the force is not proportional to the displacement from the mean position.

  Calculation of Time period:--

Time taken in traveling (l-a) distance =t.

l-a =0 +½ft² =½ft²

→t =√{2(l-a)/f}

The time period will be T =4t, because the slab travels four times (l-a) distance in a complete cycle.

So, T=4√{2(l-a)/f}

=4√[2(l-a)/{½Ԑ²εₒA(K-1)/lmd}] 

=4√[4lmd(l-a)/Ԑ²εₒA(K-1)]

=8√{lmd(l-a)/Ԑ²εₒA(K-1)}

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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

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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

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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)

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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


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

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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