Thursday, March 11, 2021

H C Verma solutions, Capacitors, EXERCISES, Q41-Q50 Chapter-31, Concepts of Physics, Part-II

Capacitors

EXERCISES, Q41 - Q50


   41. A point charge Q is placed at the origin. Find the electrostatic energy stored outside the sphere of radius R centered at the origin.  



Answer: The sphere will have charge Q appeared on the outer surface. The capacitance of a spherical capacitor is given as,

C =4πεₒR. 

Hence the energy stored in this capacitor U = Q²/2C

   = Q²/8πεₒR. 




 


   42. A metal sphere of radius R is charged to a potential V. (a) Find the electrostatic energy stored in the electric field within a concentric sphere of radius 2R. (b) Show that the electrostatic field energy stored outside the sphere of radius 2R equals that stored within it. 



Answer: As we saw in the previous problem, the electrostatic energy stored outside a spherical capacitor of radius R is,

U =Q²/(8πεₒR), 

Hence the electrostatic energy stored outside a spherical capacitor of radius 2R is,

U' =Q²/(8πεₒ*2R) =Q²/(16πεₒR)

(a) Hence the electrostatic energy stored in the electric field within a concentric sphere of radius 2R =U -U'

 =Q²/(8πεₒR) -Q²/(16πεₒR)

 =Q²/(16πεₒR) 

 =(CV)²/(16πεₒR)

{Put the value of C} 

 =(4πεₒRV)²/(16πεₒR)

 =(16π²εₒ²R²V²)/(16πεₒR)

 =πεₒRV².


(b) Electrostatic field energy stored outside the sphere of radius 2R is

U' = Q²/(16πεₒR) ---- (i)

Electrostatic field energy stored inside this sphere is the same as the energy stored between the sphere of radius R and the sphere of radius 2R because the field inside the smaller sphere is zero. This energy has been derived in the above solution (a) as  

U -U' =Q²/(8πεₒR) -Q²/(16πεₒR)

 =Q²/(16πεₒR) ------- (ii)

Comparing (i) and (ii), both are equal. 


  


 


   43. A large conducting plane has a surface charge density of 1.0x10⁻⁴ C/m². Find the electrostatic energy stored in a cubical volume of edge 1.0 cm in front of the plane. 



Answer: Given σ =1.0x10⁻⁴ C/m².

The electric field in front of the large conducting plane,

E = σ/εₒ 

Energy per unit volume in a space having field E is,

u =½εₒE²

Cubical volume = a³ where a is the edge of the cube. Hence the energy stored in the cubical volume,

U =½εₒE²a³

   =½εₒ(σ/εₒ)²a³

   =½σ²a³/εₒ 

Putting the respective values we get,

U=½*(1.0x10⁻⁴)²*(0.01)³/8.85x10⁻¹² J

 =5.6x10⁻⁴ J.

        





 

   44. A parallel-plate capacitor having a plate area 20 cm² and separation between the plates 1.00 mm is connected to a battery of 12.0 V. The plates are pulled apart to increase the separation to 2.0 mm. (a) Calculate the charge flown through the circuit during the process. (b) How much energy is absorbed by the battery during the process? (c) Calculate the stored energy in the electric field before and after the process. (d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart. (e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.  

 


Answer: Given, A =20 cm² =0.002 m².

d = 1 mm =0.001 m, V =12.0 volt, hence the capacitance,

C =εₒA/d =εₒ*0.002/0.001 

   =2εₒ

Charge on this capacitor, Q =CV

→Q =2εₒ*12 =24εₒ 

When the distance is increased to 2 mm, the capacitance is now 

C' = εₒ*0.002/0.002 =εₒ

Charge on this capacitor, Q' =C'V

→Q' =εₒ*12 =12εₒ

(a) Hence the charge flown through the circuit during the process is  

Q-Q' = 24εₒ -12εₒ =12εₒ

  =12*8.85x10⁻¹² F

  =1.06x10⁻¹⁰ C.


(b) The the energy absorbed by the battery during the process 

=(Q-Q')V

=12εₒV  

=12*8.85x10⁻¹²*12 J

=12.7x10⁻¹⁰ J

 

(c) Stored energy in the electric field before the process

U =½QV

   =½*24εₒ*12

   =144εₒ

   =144*8.85x10⁻¹² J

   =12.7x10⁻¹⁰ J

Stored energy in the electric field after the process,
U' =½Q'V

   =½*12εₒ*12

   =64εₒ

  =64*8.85x10⁻¹² J

  =6.37x10⁻¹⁰ J. 


(d) Since the capacitor is connected to the battery, the charge on it varies during the pulling process. The charge on the capacitor Q =CV =εₒAV/x, where x is the distance between the plates.

    The electric field due to the positive plate, E =σ/2εₒ =Q/2Aεₒ

The force on the negative plate having charge = -Q will be

F =(-Q)E =-QE 

The person will do work against this force in pulling the plates apart. He will have to pull it slowly and the force by him will be =QE.

Work done by him in pulling the plate by a small distance dx,

dU =QEdx =Q*(Q/2Aεₒ)dx

→dU =(Q²/2Aεₒ)dx 

→dU ={(εₒAV/x)²/2Aεₒ}dx 

→dU =[{(εₒAV²)/2}/x²]dx

Hence U =(εₒAV²/2)∫(1/x²)dx

=(εₒAV²/2)*[-1/x]

Applying lower limit x =1 mm =0.001 m, upper limit, x =2 mm =0.002 m, we get

U =(εₒAV²/2)*[-1/0.002 +1/0.001]

 =½εₒAV²*500

=250*8.85x10⁻¹²*0.002*12² J

=6.37x10⁻¹⁰ J.


(e) The energy absorbed by the battery is equal to the sum of the loss of the stored energy during the process and the work done by the person. Hence no heat is produced during the process. 



 

 

   45. A capacitor having a capacitance of 100 µF is charged to a potential difference of 24 V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12 V with the positive plate of the capacitor joined with the positive terminal of the battery. (a) Find the charge on the capacitor before and after the reconnection. (b) Find the charge flown through the 12 V battery. (c) Is work done by the battery or is it done on the battery? Find its magnitude. (d) Find the decrease in the electrostatic field energy. (e) Find the heat developed during the flow of charge after reconnection.   



Answer: (a) Given C =100 µF. Before reconnection V =24 volts. Hence the charge on the capacitor,

Q =CV =100*24 µC =2400 µC.

After reconnection, V =12 volts. Hence the charge now is

Q' =100*12 µC =1200 µC.


(b) The charge flown through the 12 V battery = Q -Q' =2400 -1200 µC

   =1200 µC.


(c) Since the charge of 1200 µC flows against the emf of 12 V battery, the work is done on the battery. Its magnitude =(Q -Q')V

  =1200*12 µJ

  =14400 µJ

  =14.4 mJ.


(d) The decrease in electrostatic field energy =U -U'

=½QV -½Q'V'

=½*2400*24 -½*1200*12 µJ

=28800 -7200 µJ

=21600 µJ

=21.6 mJ.


(e) The loss of electrostatic field energy =21.6 mJ but the work done on the battery =14.4 mJ. Hence the heat developed during the flow of charge after reconnection =21.6 -14.4 mJ =7.2 mJ.           





 

 

 

   46. Consider the situation shown in figure (31-E23). The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery. (c) Find the change in energy stored in the capacitors. (d) Find the heat developed in the system.
The figure for Q-46

    


Answer: (a) Since the switch is open for a long time both the capacitors are in series. Equivalent capacitance C' is

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

→C' =C/2

The charge on the equivalent capacitor

Q =C'Ɛ =CƐ/2

When the switch S is closed, the plates of the capacitor on the right side have both come to the same potential and the earlier charges on its plate nullify each other. Now only one capacitor on the left is effective. Charge on this capacitor,

Q' =CƐ.

Hence the charge flew through the battery when switch S is closed =Q'-Q

=CƐ -CƐ/2 =CƐ/2. 


(b) The work done by the battery

 =Charge flown*emf

 =(Q'-Q)*Ɛ =CƐ/2*Ɛ

 =CƐ²/2.


(c) Energy stored initially U =½QV

=½*(CƐ/2)*Ɛ =CƐ²/4 

Energy stored after the switch S is closed, U' =½Q'Ɛ =½CƐ*Ɛ =½CƐ²

Hence the change in energy stored in capacitors =U' -U

=½CƐ² -CƐ²/4

=CƐ²/4.


(d) The work done by the battery {from (b)} =CƐ²/2

The change in energy stored in capacitors {from (c)} =CƐ²/4

The difference is the loss of energy which is used to develop heat in the system. It is equal to,

CƐ²/2 -CƐ²/4 =CƐ²/4.       




 


   47. A capacitor of capacitance 5.00 µF is charged to 24 V and another capacitor of capacitance 6.0 µF is charged to 12.0 V. (a) Find the energy stored in each capacitor. (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitor. (c) Find the loss of electrostatic energy during the process. (d) Where does this energy go? 



Answer: (a) C =5 µF, V = 24 volts,

Energy stored, E =½CV²

→E =½*5*24² =1440 µJ =1.44 mJ.

         For the other capacitor, C' =6 µF, V' =12 volts, 

hence the energy stored, E' =½C'V'²

→E' =½*6*12² =432 µJ =0.432 mJ.


(b) The charge Q on the first capacitor,

Q =CV =5*24 =120 µC

Charge on the second capacitor,

Q' =C'V' =6*12 =72 µC

When the 120 µC plate of the first capacitor is connected to the -72 µC plate of the other, the net total charge on both the connected plate Q" =120-72 =48 µC. Similarly, the net total charge on another pair of connected plates =-120 +72 =-48 µC. 

Diagram for the Q-47


   Let V" be the potential difference across the plates. Then,

Q" =C"V"

We can see that the plates are connected in parallel. Hence , 

C" =C +C' = 5+6 =11 µF.

V" =Q"/C" =48/11 volts

Now the charge on the first capacitor

q =CV" =5*48/11 =21.8 µC.

The charge on the second capacitor,

q' =C'*V" =6*48/11 =26.2 µC.


(c) The electrostatic energy before connection {from (a)}, 

U =E+E' =1.44+0.432 mJ =1.872 mJ

The ectrostatic energy after connection,

U' =½C"V"² =½*11*(48/11)² µJ

    =104.73 µJ

    =0.105 mJ

Hence the loss of electrostatic energy during the process =U -U'

            =1.872 -0.105 mJ

            ≈1.77 mJ.


(d) This lost energy is dissipated as heat.               




  


   48. A 5.0 µF capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.    



Answer: C = 5 µF, V =12 volts. The charge on the capacitor, Q =CV 

→Q =5*12 =60 µC.

So the charge on the positive plate of the capacitor is 60 µC and on the negative plate -60 µC.

When the plates of this capacitor are connected to a 12 V battery with reverse polarity, the negative plate becomes positive and the positive plates become negative. Since the potential difference is still 12 V, the charge on the capacitor is 60 µC. Consider the initial negative plate with charge -Q. After connection to the battery, this plate has now charge Q. Hence the charge supplied by the battery after connection =Q-(-Q) =2Q. Hence the energy supplied by the battery U =(2Q)V,

→U =2*60*12 µJ =1440 µJ.

Since the charge on the capacitor and the potential difference across it remains the same before and after the connection with the battery, the energy stored in it remains the same. Hence the total energy supplied by the battery is used in heating the wires. So the heat developed in the connecting wires =1440 µJ =1.44 mJ.            





 

   49. The two square faces of of a rectangular dielectric slab (dielectric constant 4.0) of dimensions 20 cmx20 cmx1.0 mm are metal coated. Find the capacitance between the coated surfaces.   



Answer: Dielectric constant of the slab, K = 4.0, A =20 cm x 20 cm =400 cm² =0.04 m², d =1.0 mm =0.001 m.

The capacitance between the coated surfaces, C =KεₒA/d

→C =4.0*8.85x10⁻¹²*0.04/0.001 F

→C =1.42x10⁻⁹ F =1.42 nF.   




 


   50. If the above capacitor is connected across a 6.00 V battery, find (a) the charge supplied by the battery, (b) the induced charge on the dielectric and (c) the net charge appearing on one of the coated surfaces.    



Answer: The potential difference across this capacitor =6.00 V, C =1.42 nF.

(a) The charge supplied by the battery,

Q =CV =1.42x6.00 nC =8.52 nC.


(b) The induced charge on the dielectric,

Qp =Q(1 -1/K)

     =8.52*(1-1/4) nC

     =8.52*3/4 nC

     =6.4 nC. 


(c) The net charge appearing on one of the coated surfaces =Q -Qp

  =8.5 -6.4 =2.1 nC.        

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EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

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CHAPTER- 24-Kinetic Theory of Gases







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CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


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CHAPTER- 11 - Gravitation




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