Wednesday, March 17, 2021

H C Verma solutions, Capacitors, EXERCISES, Q51-Q60 Chapter-31, Concepts of Physics, Part-II

Capacitors

EXERCISES, Q51 - Q60


   51. The separation between the plates of a parallel-plate capacitor is 0.500 cm and its plate area is 100 cm². A 0.400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.   



Answer: When the metal plate is inserted into the parallel plate capacitor, both of its parallel surfaces get induced charges. Two gaps on either side make two capacitors joined by the metal in series.
Diagram for Q-51

  Let the gap on one side = x, plate thickness = t, separation of the given capacitor =d. So the gap on the other side =d-t-x.

   The capacitance of the capacitor having gap =x,

  C' =εₒA/x, where A is the plate area.

    The capacitance of the capacitor having gap d-t-x,

     C" =εₒA/(d-t-x)

Equivalent capacitance C  is given as

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

→C =C'C"/(C' +C")

 =(εₒA)²/[x(d-t-x)(εₒA){1/x +1/(d-t-x)}]

 =εₒA/[x(d-t-x)*(d-t-x+x)/{x(d-t-x)}]

 =εₒA/(d-t). 

This expression for the capacitance of the assembly is independent of x, i.e. independent of the position of the metal plate within the gap. We can find the value of C by putting the given data.

Given, d =0.500 cm =0.005 m,

t =0.400 cm =0.004 m,

A =100 cm² =0.01 m²  

C =εₒA/(d-t)

  =8.85x10⁻¹²*(0.01)/(0.005-0.004)

 =8.85x10⁻¹²*10 F

 =88.5x10⁻¹² F

 =88.5 pF. 

       




 

  52. A capacitor stores 50 µC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 µC flows through the battery. Find the dielectric constant of the material inserted. 



Answer: Initial charge, Q =50 µC.

When th gap is filled with dielectric having dielectric constant K, 100 µC flows through the battery. So now the charge, Q' =50 +100 =150 µC. 

Initially, capacitance C =εₒA/d

Finally, capacitance C' =KεₒA/d

Hence C'/C =K

→(Q'/V)/(Q/V) =K, {V=Battery emf}

→Q'/Q =K

→K =Q'/Q =150/50 =3.         



 


   53. A parallel plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?



Answer: (a) d = 2 mm =0.002 m.

C =5 µF, V = 6 volts. Since Q =CV,

The charge on the capacitor

Q =(5 µF)*(6 V) =30 µC.

It means the charge on the positive plate of the capacitor is +30 µC.


(b) The electric field between the plates,

E = V/d

   =(6 V)/(0.002 m)

   =3000 V/m

   =3x10³ V/m.


(c) Initial Capacitance, C =εₒA/d

→5x10⁻⁶ =εₒA/d

→εₒA =5x10⁻⁶*d -------------(i)

Thickness of dielectric = t = 1 mm =0.001 m

Dielectric constant = K =5

Now the capacitance of the combination

C* =εₒA/{d-t(1-1/K)}

Putting value from (i)

  =5x10⁻⁶*d/{d-t(1-1/k)}

  =5x10⁻⁶*2/{2-1(1-1/5)}

{Since numerator and denominator have d and t, so we have not to change the unit} 

 =5x10⁻⁶*2/(6/5)

  =8.33 µF. 


(d) The charge flown through the battery after inserting the dielectric =Q' -Q

=C*V -CV

=(C* -C)V

=(8.33 -5)*6

=3.33*6 µC

=20 µC.          




 

   54. A parallel plate capacitor has a plate area 100 cm² and plate separation 1.0 cm. A glass plate (dielectric constant 6.0) of thickness 6.0 mm and an ebonite plate (dielectric constant 4.0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.  



Answer: A = 100 cm² =0.01 m².

The two dielectrics will act as separate capacitors connected in series, the junction of dielectrics being a common plate for the two.

Separation of the first capacitor, d = 6 mm, Dielectric constant K = 6.0 and separation of the second d' =4.0 mm and K' =4.0,

C =εₒAK/d, and

C' =εₒAK'/d'.

The capacitance of the assembly C͓ is given as,

1/C͓ =1/C +1/C'

→C͓ =C'C/(C+C')

Now, C'C =(εₒAK'/d')*(εₒAK/d) 

         =(εₒA)²K'K/d'd

And C +C' =εₒA{K'/d' +K/d}

       =εₒA(K'd +Kd')/d'd

Hence 

C͓ ={(εₒA)²K'K/d'd}/{εₒA(K'd+Kd')/d'd}

=εₒAK'K/(K'd+Kd') 

=8.85x10⁻¹²*0.01*6*4/(4*0.006+6*0.004)

=44.25x10⁻¹² F

≈44 pF. 




 

   55. A parallel plate capacitor having a plate area 400 cm² and separation between the plates 1.00 mm is connected to the power supply of 100 V. A dielectric slab of thickness 0.5 mm and dielectric constant 5.0 is inserted into the gap. (a) Find the increase in electrostatic energy. (b) If the power supply is now disconnected and the dielectric slab is taken out, find a further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out? 



Answer: A =400 cm² =0.04 m², d =1 mm =0.001 m, V =100 volts. K =5.

Capacitance, C =εₒA/d

=8.85x10⁻¹²*0.04/0.001 F

=3.54x10⁻¹⁰ F

Stored electrostatic energy,

U = ½CV²

   =½*3.54x10⁻¹⁰*10⁴

   =1.77x10⁻⁶ J 

   =1.77 µJ

After the dielectric slab is inserted, Capacitance 

C' =εₒA/{d-t(1-1/K)}

=8.85x10⁻¹²*0.04/(0.001-0.0005*0.8)

=5.90x10⁻¹⁰ F

Stored energy now,

U' =½C'V²

 =½*5.9x10⁻¹⁰*10⁴

 =2.95x10⁻⁶ J

 =2.95 µJ

Increase in stored energy =U'-U

=2.95 -1.77 µJ

=1.18 µJ.


(b) In the final state charge on the capacitor, Q' =C'V

→Q' =5.90x10⁻¹⁰*100 C

   =5.90x10⁻⁸ C

When the power supply is disconnected this charge remains on the capacitor. When the dielectric is also removed its capacitance goes back to original capacitance C. Now the energy =Q'²/2C

=(5.90x10⁻⁸)²/(2*3.54x10⁻¹⁰)

=4.92 µJ 

Hence the further increase in the energy =4.92 -2.95 µJ

=1.97 µJ.


(c) When the dielectric slab is inserted, the capacitance increases and the capacitor draws more charge from the battery. Hence the energy stored increases in this process.

   In the second case, the battery is disconnected, the increased charge remains on the capacitor. But with the withdrawal of dielectric, the capacitance reduces and reverts back to the original one. Hence in the expression of stored energy, U =Q²/2C, since C <C', U increases. 

  



 

   56. Find the capacitances of the capacitors shown in figure (31-E24). The plate area is A and the separation between the plates is d. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.  
The figure for Q -56


Answer: (a) In this part, the assembly is equivalent to two capacitors joined in series. Separation for each capacitor =d/2.

For the first, C' =εₒAK₁/(d/2)

           =2εₒAK₁/d

For the second , C" =εₒAK₂/(d/2)

           =2εₒAK₂/d

Hence the capacitance of the assembly,

 C =C'C"/(C'+C")

 =(2Aεₒ/d)²K₁K₂/{(2εₒA/d)(K₁+K₂)}

 =2εₒAK₁K₂/{d(K₁+K₂)}.

 

(b) In the second part the situation is similar. Just the difference is that as if three capacitors are are connected in series and the separation of each capacitor is d/3. Capacitances of them are,

First, C' =εₒAK₁/(d/3)

            =3εₒAK₁/d

Similarly for the second

       C" =3εₒAK₂/d

For third, C‴ =3εₒAK₃/d

The equivalent capacitance C is,

1/C =1/C' +1/C" +1/C‴

   =(C'C"+C"C‴+C‴C')/C'C"C‴

→C =C'C"C‴/(C'C"+C"C‴+C‴C')

   =(3εₒA/d)³K₁K₂K₃/{(3εₒA/d)²(K₁K₂+K₂K₃+K₃K₁)}

=3εₒAK₁K₂K₃/{d(K₁K₂+K₂K₃+K₃K₁)}


(c) In this part two capacitors with dielectrics are connected in parallel. The separation of each is d but the plate area for each of them =A/2.

C' =εₒAK₁/2d

C" =εₒAK₂/2d

In parallel, equivalent capacitance

C =C'+C"

  =(εₒA/2d)(K₁+K₂)

  =εₒA(K₁+K₂)/2d.              




  

 


   57. A capacitor is formed by two square metal plates of edge a, separated by a distance d. Dielectrics of dielectric constant K₁ and K₂ are filled in the gap as shown in figure (31-E25). Find the capacitance.  
The figure for Q -57



Answer: Since the thickness of dielectric varies, we will have to take the help of integration. Consider an infinitesimal width dx at a distance x from left as shown in the diagram. This section has two very small capacitors with plate area =adx, separation of lower capacitor =x.tanß and the separation of upper capacitor =d-x.tanß
Diagram for Q-57

These two small capacitors are connected in series. Suppose the capacitance of the upper one is dC₁ and of the lower one is dC₂. Then,

dC₁ =ε₀(adx)K₁/(d-xtanß)

dC₂ =ε₀(adx)K₂/xtanß         

The equivalent capacitance dC is given as,

1/dC =1/dC₁+1/dC₂

→dC =dC₁.dC₂/(dC₁+dC₂)

=(εₒadx)²K₁K₂/{x.tanß(d-x.tanß)}(εₒadx){K₁/(d-xtanß)+K₂/(x.tanß}

=εₒaK₁K₂.dx/{K₁xtanß+K₂(d-xtanß)]

=εₒaK₁K₂.dx/{K₂d+(K₁-K₂)tanß*x]

On integration,

∫dC={εₒaK₁K₂/(K₁-K₂)tanß}*[logₑ{K₂d+(K₁-K₂)tanß*x}]

{The integral of the underlined portion is in the form of 1/(a+bx), and we integrate it from x=0 to x=a}

→C={εₒaK₁K₂/(K₁-K₂)tanß}}{logₑ(K₂d+(K₁-K₂)a.tanß)-logₑ(K₂d)}

→C={εₒaK₁K₂/(K₁-K₂)tanß}{logₑ(K₂d+(K₁-K₂)d)/(K₂d)}

{We replace a.tanß =d}

={(εₒa²K₁K₂)/(K₁-K₂)a.tanß}*logₑ{K₁d/(K₂d)}

{Multiplying numerator and denominator by a}

={(εₒa²K₁K₂)/(K₁-K₂)d}logₑ(K₁/K₂)}

={εₒa²K₁K₂*logₑ(K₁/K₂)}/{(K₁-K₂)d}

This is the value of given composite capacitor.

 




  



   58. Figure (31-E26) shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plate of the capacitors is filled with a dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.  
The figure for Q -58



Answer: When the switch is closed, the capacitors are connected in a parallel combination. Equivalent capacitance

C' =2C

The energy stored in the capacitors,

U =½(2C)V² =CV².

When the switch is opened and the free space filled with dielectric with K=3, the capacitance of the capacitor A =KC =3C. The energy stored in capacitor A,

Uₐ =½(3C)V² =3CV²/2

Since the switch is opened, the charge on it remains the same i.e. Q=CV but capacitance increased to 3C. Hence the energy stored in it 
Uᵦ =½Q²/(3C)
    =½(C²V²)/(3C)
    =CV²/6
Hence the total energy stored in this case =U' =Uₐ+Uᵦ
→U' =3CV²/2 +CV²/6
       =10CV²/6
       =5CV²/3
Hence U/U' = CV²/(5CV²/3) =3/5
So, U : U' = 3 : 5.          

        



 


   59. A parallel plate capacitor of plate area A and the plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab. 



Answer: The capacitance of the capacitor, C =ε₀A/d. When charged with battery, Q =CV =εₒAV/d.

Energy stored U =½QV

→U =½εₒAV²/d

When the battery is disconnected, the charge remains the same i.e. = Q. When the dielectric of K is inserted, capacitance is now C' = εₒAK/d.

Hence the energy stored now is,

U' =½Q²/C'

  =½(εₒAV/d)²/(εₒAK/d)

  =½εₒAV²/Kd

Increase in energy stored =U' -U

=½εₒAV²/Kd -½εₒAV²/d 

=(½εₒAV²/d)*(1/K-1)             

The increase in energy of the system is equal to the work done on the system in the process of inserting the slab. So the work done is,

W =(½εₒAV²/d)*(1/K-1)

    =εₒAV²(1/K-1)/2d.






 

   60. A capacitor having a capacitance of 100 µF is charged to a potential difference of 50 V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates. (c) What charge would have produced this potential difference in absence of the dielectric slab? (d) Find the charge induced at the surface of the dielectric slab.



Answer: (a) The capacitance, C =100 µF, V = 50 volts. Hence the charge,

Q = CV =100*50 µC =5000 µC 

→Q =5 mC.

So the magnitude of the charge on each plate = 5 mC.


(b) When the battery is disconnected, this charge remains on the capacitor. The new capacitance of the capacitor with a dielectric of K =2.5,

C' =KC =2.5*100 =250 µF.

Q = 5000 µC.

Hence the new potential difference between the plates,

V' =Q/C'

   =5000/250 V

   =20 V.


(c) In absence of the dielectric slab, the capacitance C = 100µF. Potential difference = 20 V. The charge required to produce this potential difference,

Q =CV' =100*20 =2000 µC =2 mC.


(d) The charge induced at the surface of the dielectric slab,

Qₚ =Q{1 -1/K}

[Q =5 mC, and K =2.5]

    =5*{1 -1/2.5} mC

    =5*1.5/2.5 mC

    =5*3/5 mC

    =3 mC.                

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EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


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CHAPTER- 24-Kinetic Theory of Gases







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CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



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CHAPTER- 11 - Gravitation




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