Irodov Solutions: Electric Capacitance Energy of an Electric Field - 2

# Irodov Solutions: Electric Capacitance Energy of an Electric Field - 2 | Physics Class 12 - NEET PDF Download

Q. 117. A circuit has a section AB shown in Fig. 3.21. The emf of the source equals  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.

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

Hence, voltage across the capacitor C1

and voltage across the capacitor, C2

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

Solution. 118.  then using  in the closed circuit, (Fig.)

or,

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

and similarly

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

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

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?

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

-Δφ = 0 yields

and in the circuit 13461,

Now

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

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?

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

as they are in series combination (Fig.-a)

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

(1)

And in the closed loop BCFAB from - Δφ = 0

(2)

From (1) and (2) q1 - 0

Now, charge flown through section

and charge flown through section

Q. 123. In the circuit shown in Fig. 3.25 the emf of each battery is equal to  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.

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

and when the switch is closed,

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

through section

through the section

and through the section

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

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

In the loop 12341, using - Δφ = 0

(1)

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

(2)

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

Hence,

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.

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

And using the symmetry

and

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

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

Now, capacitance of the network,

From Eqs. (1) and (2)

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.

Solution. 127. (a) Interaction energy of any two point charges q1 and q2 is given by   is the separation between the charges.

Hence, interaction energy of the system,

and

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,

(1)

But

and putting    (2)

From Eqs. (1) and (2),

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.

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

Hence finally integrating

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 :

Now, when the capacitors are connected in parallel, equivalent capacitance of the system,   and hence, energy stored in the system :

as charge remains conserved during the process.

So, increment in the energy,

The document Irodov Solutions: Electric Capacitance Energy of an Electric Field - 2 | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
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## Physics Class 12

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## FAQs on Irodov Solutions: Electric Capacitance Energy of an Electric Field - 2 - Physics Class 12 - NEET

 1. What is electric capacitance and how is it related to energy of an electric field?
Ans. Electric capacitance is a measure of an object's ability to store electric charge. It is defined as the ratio of the magnitude of the charge stored on an object to the magnitude of the potential difference across the object. The energy of an electric field is directly related to the capacitance, as the energy stored in the electric field is given by the formula E = (1/2)CV^2, where E is the energy, C is the capacitance, and V is the potential difference.
 2. How can the energy of an electric field be calculated using capacitance?
Ans. The energy of an electric field can be calculated using the formula E = (1/2)CV^2, where E is the energy, C is the capacitance, and V is the potential difference across the object. By knowing the capacitance and the potential difference, one can easily calculate the energy stored in the electric field.
 3. What factors affect the capacitance of an object?
Ans. The capacitance of an object depends on several factors. The most important factor is the geometry of the object, specifically the shape and size of the conductive plates or surfaces involved. The distance between the plates or surfaces also affects the capacitance, with a smaller distance resulting in a higher capacitance. Additionally, the dielectric material between the plates or surfaces, if present, can also influence the capacitance.
 4. How does the energy stored in an electric field change when the capacitance is increased?
Ans. When the capacitance of an object is increased, the energy stored in the electric field also increases. This is because the energy is directly proportional to the capacitance. Therefore, a larger capacitance will result in a larger amount of energy stored in the electric field for the same potential difference.
 5. Can the energy of an electric field be negative?
Ans. No, the energy of an electric field cannot be negative. The energy is always positive or zero. Negative energy would imply that the electric field is releasing energy, which violates the conservation of energy principle. Therefore, the energy of an electric field is always considered to be positive.

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