Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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The document Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 117. A circuit has a section AB shown in Fig. 3.21. The emf of the source equals Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev the capacitor capacitances are equal to C1 =  1.0 μF and C2 = 2.0 μF, and the potential difference φA - φB = 5.0 V. Find the voltage across each capacitor. 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 117. Let, us make the charge distribution, as shown in the figure.

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev
Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Hence, voltage across the capacitor C1

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and voltage across the capacitor, C2

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 118. In a circuit shown in Fig. 3.22 find the potential difference between the left and right plates of each capacitor. 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 118. Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev then using Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev in the closed circuit, (Fig.)

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

or,   Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Hence the P.D. accross the left and right plates of capacitors,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and similarly

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 119. Find the charge of each capacitor in the circuit shown in Fig. 3.22. 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 119. Taking benefit of tBe foregoing problem, the amount of charge on each capacitor

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 120. Determine the potential difference φ- φB between points A and B of the circuit shown in Fig. 3.23. Under what condition is it equal to zero? 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 120. Make the charge distribution, as shown in the figure. In the circuit, 12561.

 -Δφ = 0 yields

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and in the circuit 13461,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Now  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 121. A capacitor of capacitance C1  = 1.0 μF charged up to a voltage V = 110 V is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing the capacitances C2 = 2.0 μF and C3 = 3.0 μR. What charge will flow through the connecting wires?

Solution. 121. Let, the charge q flows through the connecting wires, then at the state of equilibrium, charge distribution will be as shown in the Fig. In the closed circuit 12341, using

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 122. What charges will flow after the shorting of the switch Sw in the circuit illustrated in Fig. 3.24 through sections 1 and 2 in the directions indicated by the arrows? 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 122. Initially, charge on the capacitor C1 or C2,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev as they are in series combination (Fig.-a)

when the switch is closed, in the circuit CDEFC from - Δφ = 0, /(Fig. b) 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev    (1)

And in the closed loop BCFAB from - Δφ = 0 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev   (2)

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRevIrodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

From (1) and (2) q1 - 0 

Now, charge flown through section  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and charge flown through section  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 123. In the circuit shown in Fig. 3.25 the emf of each battery is equal to Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev and the capacitor capacitances are equal to C1  = 2.0 μF and C2 = 3.0 μF. Find the charges which will flow after the shorting of the switch Sw through sections' 1, 2 and 3 in the directions indicated by the arrows.

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 123. When the switch is open, (Fig-a)

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRevIrodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and when the switch is closed,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Hence, the flow of charge, due to the shortening of switch,

through section  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

through the section  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and through the section   Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 124. Find the potential difference φA — φB  between points A and B of the circuit shown in Fig. 3.26.

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 124. First of all, make the charge distribution, as shown in the figure. 

In the loop 12341, using - Δφ = 0

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev    (1)

Similarly, in the loop 61456, using - Δφ = 0

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev   (2)

From Eqs. (1) and (2) we have

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Hence,   Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 125. Determine the potential at point 1 of the circuit shown in Fig. 3.27, assuming the potential at the point O to be equal to zero. Using the symmetry of the formula obtained, write the expressions for the potentials. at points 2 and 3. 

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 125. In the loop ABDEA, using - Δφ = 0

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

And using the symmetry  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

The answers have wrong sign in the book.


Q. 126. Find the capacitance of the circuit shown in Fig. 3.28 between points A and B.

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 126. Taking the advantage of symmetry of the problem charge distribution may be made, as shown in the figure.
In the loop, 12561, - Δφ = 0

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Now, capacitance of the network,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

From Eqs. (1) and (2)

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 127. Determine the interaction energy of the point charges located at the corners of a square with the side a in the circuits shown in Fig. 3.29.

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Solution. 127. (a) Interaction energy of any two point charges q1 and q2 is given by  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev is the separation between the charges.

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRevIrodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Hence, interaction energy of the system,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev
and Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 128. There is an infinite straight chain of alternating charges q and -q. The distance between the neighbouring charges is equal to a. Find the interaction energy of each charge with all the others.
 Instruction. Make use of the expansion of In (1 + α) in a power series in α.

Solution. 128. As the chain is of infinite length any two charge of same sign will occur symmetrically to any other charge of opposite sign.

So, interaction energy of each charge with all the others,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev    (1)

But    Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

and putting  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev  (2)

From Eqs. (1) and (2),

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 129. A point charge q is located at a distance l from an infinite tonducting plane. Find the interaction energy of that charge with chose induced on the plane. 

Solution. 129. Using electrical im age method, interaction energy of the charge q with those induced on the plane.

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 130. Calculate the interaction energy of two balls whose charges qi  and q, are spherically symmetrical. The distance between the centres of the balls is equal to l.
 Instruction. Start with finding the interaction energy of a ball and a thin spherical layer.

Solution. 130. Consider the interaction energy of one of the balls (say 1) and thin spherical shell of the other. This interaction energy can be written as  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRevIrodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev
Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Hence finally integrating

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev


Q. 131. A capacitor of capacitance C1 = 1.0 μF carrying initially a voltage V = 300 V is connected in parallel with an uncharged capacitor of capacitance C2 = 2.0 μF. Find the increment of the electric energy of this system by the moment equilibrium is reached. Explain the result obtained.

Solution. 131. Charge contained in the capacitor of capacitance C1 is q = C1 φ and the energy, stored in it :

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

Now, when the capacitors are connected in parallel, equivalent capacitance of the system,  Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev and hence, energy stored in the system :

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev  as charge remains conserved during the process. 

So, increment in the energy,

Irodov Solutions: Electric Capacitance Energy of An Electric Field- 2 Notes | EduRev

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