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Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET PDF Download

Q. 101. Find the capacitance of an isolated ball-shaped conductor of radius R1 surrounded by an adjacent concentric layer of dielectric with permittivity ε and outside radius R2.

Solution. 101. Let us mentally impart a charge q on the conductor, then

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Hence the sought capacitance,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 102. Two parallel-plate air capacitors, each of capacitance C, were connected in series to a battery with Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET Then one of the capacitors was filled up with uniform dielectric with permittivity ε. How many times did the electric field strength in that capacitor decrease? What amount of charge flows through the battery?

Solution. 102. From the symmetry o f the problem, the voltage across each capacitor, Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET and charge on each capacitor Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET in the absence of dielectric.
Now when the dielectric is filled up in one of the capacitors, the equivalent capacitance of the system,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

and the potential difference across the capacitor, which is filled with dielectric,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

But  φα E

So, as φ decreases Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET times, the field strength also decreases by the same factor and flow of charge Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 103. The space between the plates of a parallel-plate capacitor is filled consecutively with two dielectric layers 1 and 2 having the thicknesses d1 and d2 and the permittivities ε1 and ε2  respectively. The area of each plate is equal to S. Find:
 (a) the capacitance of the capacitor;
 (b) the density σ' of the bound charges on the boundary plane if the voltage across the capacitor equals V and the electric field is directed from layer 1 to layer 2.

Solution. 103. (a) As it is series combination of two capacitors,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

(b) Let, a be the initial surface charge density, then density of bound charge on the boundary plane.

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

But,  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

So,  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 104. The gap between the plates of a parallel-plate capacitor is filled with isotropic dielectric whose permittivity ε varies linearly from ε1 to ε22 > ε1) in the direction perpendicular to the plates. The area of each plate equals S, the separation between the plates is equal to d. Find: 
 (a) the capacitance of the capacitor; 
 (b) the space density of the bound charges as a function of ε if the charge of the capacitor is q and the field E in it is directed toward the growing ε values. 

Solution. 104. (a) We point the jt-axis lowards right and place the origin on the left hand side plate. The left plate is assumed to be positively charged.

Since e varies linearly, we can write,

ε(x) = a + bx

where a and b can be determined from the boundary condition. We have 

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Thus,  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Now potential difference between the plates 

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Hence, the sought capacitance Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

and the space density of bound charges is

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 105. Find the capacitance of a spherical capacitor whose electrodes have radii R1 and R2 > R1 and which is filled with isotropic dielectric whose permittivity varies as ε = a/r, where a is a constant, and r is the distance from the centre of the capacitor. 

Solution. 105. Let, us mentally impart a charge q to the conductor. Now potential difference between the plates,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Hence, the sought capacitance,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 106. A cylindrical capacitor is filled with two cylindrical layers of dielectric with permittivities ε1 and ε2. The inside radii of the layers are equal to R1 and R2 > R1. The maximum permissible values of electric field strength are equal to E1m and E2m  for these dielectrics. At what relationship between ε, R, and Em, will the voltage increase result in the field strength reaching the breakdown value for both dielectrics simultaneously? 

Solution. 106. Let λ be the linear chaige density then,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET     (1)
and,   Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET     (2)

The breakdown in either case will occur at the smaller value of r for a simultaneous breakdown of both dielectrics.

From (1) and (2)

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET which & the sought relationship.


Q. 107. There is a double-layer cylindrical capacitor whose parameters are shown in Fig. 3.16. The breakdown field strength values for these dielectrics are equal to E1 and E2 respectively. What is the breakdown voltage of this capacitor if ε1R1E1< ε2R2E2

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Solution. 107. Let, λ be the linear chaige density then, the sought potential difference,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

is the maximum acceptable value, and for values greater than  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET dielectric breakdown will take place,

Hence, the maximum potential difference between the plates,   

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEETIrodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 108. Two long straight wires with equal cross-sectional radii a are located parallel to each other in air. The distance between their axes equals b. Find the mutual capacitance of the wires per unit length under the condition b ≫ a.

Solution. 108. Let us suppose that linear chaige density of the wires be λ then, the potential difference, Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET The intensity of the electric field created by one of the wires at a distance x from its axis can be easily found with the help of the Gauss’s theorem,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Henee, capacitance, per unit length,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 109. A long straight wire is located parallel to an infinite conducting plate. The wire cross-sectional radius is equal to a, the distance between the axis of the wire and the plane equals b. Find the mutual capacitance of this system per unit length of the wire under the condition a ≪ b. 

Solution. 109. The field in the region between the conducting plane and the wire can bt obtained by using an oppositely charged wire as an image on the other side.
Then the potential difference between the wire and the plane,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Hence, the sought mutual capacitance of the system per unit length of the wire Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 110. Find the capacitance of a system of two identical metal balls of radius a if the distance between their centres is equal to b, with b ≫ a. The system is located in a uniform dielectric with permittivity a. 

Solution. 110. When b >> a, the charge distribution on each spherical conductor is practically unaffected by the presence of the other conductor. Then, the potential φ+-) on the positive (respectively negative) charged conductor is 

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Note : if we require tetms which depend on a/b, we have to take account of distribution of charge on the conductors.


Q. 111. Determine the capacitance of a system consisting of a metal ball of radius a and an infinite conducting plane separated from the centre of the ball by the distance l if l ≫ a. 

Solution. 111.  As in Q.109 we apply the method of image. Then the potentical difference between the +vely charged sphere and the conducting plane is one half the nominal potential difference between the sphere and its image and is

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Thus

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 112. Find the capacitance of a system of identical capacitors between points A and B shown in (a) Fig. 3.17a; (b) Fig. 3.17b. 

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Solution. 112.

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEETIrodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

(a)  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

The arrangement of capacitors shown in the problem is equivalent to the arrangement shown in the Fig.

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEETIrodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

and hence the capacitance between A and B is,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

(B) From the symmetry of the problem, there is no P.d. between D and E.. So, the combination reduces to a simple arrangement shown in the Fig and hence the net capacitance,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 113. Four identical metal plates are located in air at equal distances d from one another. The area of each plate is equal to S. Find the capacitance of the system between points A and B if the plates are interconnected as shown (a) in Fig. 3.18a; (b) in Fig. 3.18b. 

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Solution. 113. (a) In the given arrangement, we have three capacitors of equal capacitance  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET and the first and third plates are at the same potential.

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Hence, we can resolve the network into a simple form using series and parallel grouping of capacitors, as shown in the figure. Thus the equivalent capacitance

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

(b) Let us mentally impart the charges +q and -q to the plates 1 and 2 and then distribute them to other plates using charge conservation and electric induction. (Fig.). As the potential difference between the plates 1 and 2 is zero,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

The potential difference between A and B,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Hence the sought capacitane,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 114. A capacitor of capacitance C1 = 1.0 μ.F withstands the maximum voltage V1 =  6.0 kV while a capacitor of capacitance C2 = 2.0 μF, the maximum voltage V2 = 4.0 kV. What voltage will the system of these two capacitors withstand if they are connected in series? 

Solution. 114. Amount of charge, that the capacitor of capacitance C1 can withstand, q1 = C1 V1 and similarly the charge, that the capacitor of capacitance C2 can withstand, q2 = C2 V2. But in series combination, charge on both the capacitors will be same, so, qmax, that the combination can withstand = C1V1,

as C1 V1 < C2 V2, from the numerical data, given.

Now, net capacitance of the system,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

and hence,  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 115. Find the potential difference between points A and B of the system shown in Fig. 3.19 if the emf is equalIrodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET = 110 V and the capacitance ratio C2/C1 = η = 2.0.

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Solution. 115. Let us distribute the charges, as shown in the figure.

Now, we know that in a closed circuit, - Δφ = 0

So, in the loop, DCFED, 

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET    (1)

Again in the loop DGHED,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET    (2)

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Using Eqs. (1) and (2), we get

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Now,  Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

or,   Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET


Q. 116. Find the capacitance of an infinite circuit formed by the repetition of the same link consisting of two identical capacitors, each with capacitance C (Fig. 3.20). 

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Solution. 116. The infinite circuit, may be reduced to the circuit, shown in the Fig. where, C0 is the net capacitance of the combination.

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET

we get,

Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET taking only +ve value as C0 can not be negative.

The document Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
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FAQs on Irodov Solutions: Electric Capacitance Energy of an Electric Field - 1 - Physics Class 12 - NEET

1. What is electric capacitance and how is it related to the energy of an electric field?
Ans. Electric capacitance is a measure of the ability of a system to store an electric charge. It is defined as the ratio of the magnitude of the charge stored on one of the conductors to the potential difference between them. The energy of an electric field is directly related to the capacitance of the system. The energy stored in an electric field is given by the equation: Energy = (1/2) * capacitance * potential difference squared.
2. How can the energy stored in an electric field be calculated?
Ans. The energy stored in an electric field can be calculated using the equation: Energy = (1/2) * capacitance * potential difference squared. First, determine the capacitance of the system, which is the ratio of the charge stored on one of the conductors to the potential difference between them. Then, square the potential difference and multiply it by half of the capacitance to obtain the energy stored in the electric field.
3. Can the energy stored in an electric field be negative?
Ans. No, the energy stored in an electric field cannot be negative. The energy stored in an electric field is always positive because it represents the work done in moving charges against the electric field. Negative energy would imply that work is being done by the electric field to move the charges, which is not possible.
4. How does the energy stored in an electric field change with the capacitance and potential difference?
Ans. The energy stored in an electric field is directly proportional to both the capacitance and the square of the potential difference. Increasing the capacitance or the potential difference will result in an increase in the energy stored. Similarly, decreasing the capacitance or the potential difference will lead to a decrease in the energy stored.
5. What are some real-life applications of the energy stored in an electric field?
Ans. The energy stored in an electric field has various practical applications. It is used in electric power systems to store energy in capacitors, which can be later released for various purposes. Capacitors are commonly used in electronic devices to store and release energy quickly. They are also used in flash photography, defibrillators, and energy storage systems for renewable energy sources like solar and wind power.
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