A capacitor is a system of two conductors ...

CBSE Class 12 Physics 2023 Delhi Set 1 Paper

Question : 35 of 35

Marks: +1, -0

A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain desired value of capacitance.

(i) Find the equivalent capacitance between points

A

and

B

in the given diagram.

(ii) A dielectric slab is inserted between the plates of a parallel plate capacitor. The electric field between the plates decreases. Explain.

(iii) A capacitor A of capacitance

C

, having charge

Q

is connected across another uncharged capacitor

B

of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.

(iii) Two slabs of dielectric constants

2 K

and

K

fill the space between the plates of a parallel plate capacitor of plate area

bo A

and plate separation

d

as shown in figure. Find an expression for capacitance of the system.

Solution: 👈: Video Solution

Ans. (i) The given combination:

Initially the capacitor between A and B is not considered.

So, the circuit may be redrawn as

This is like a balanced Wheatstone bridge. So, the capacitor between

E

and

F

may not be considered. So, the circuit is further modified as

The capacitance between

A

and

B

C

Now capacitor between A and B (left out initially) is considered.

So, the equivalent capacitance between

\mathrm{A}

and

B

2 C

(ii) When a dielectric is placed between charged plates of a capacitor, the polarization of dielectric produces an electric field opposing the field produced by the charges on the plate. Thus, the resultant electric field between the plates decreases.

(iii) Both capacitors will attain a common potential,

V_C

(a)

V_C\;=\;\frac{\;\text{ Total charge }\;}{\;\text{ Total capacitance }\;}

\;=\;\frac{Q}{C+2C}=\;\frac{Q}{3C}

(b) Final charge on capacitor

A

\;=Q_A=C V_C

\;=\;\frac{C Q}{3 C}=\;\frac{Q}{3}

So, charge lost by capacitor

A

=Q-\;\frac{Q}{3}=\;\frac{2 Q}{3}

By placing the dielectric slab, it behaves like two capacitors in series

C_1\;=\;\frac{K_1 \varepsilon_0 A}{d/3}=\;\frac{6 K \varepsilon_0 A}{d}

C_2\;=\;\frac{K_2 \varepsilon_0 A}{2d/3}=\;\frac{3 K \varepsilon_0 A}{2d}

\;\frac{1}{C_{\;\text{eq}\;}}\;=\;\frac{1}{C_1}+\;\frac{1}{C_2}

Or,

\;\frac{1}{C_{\;\text{eq}\;}}\;=\;\frac{d}{6 K \varepsilon_0 A}+\;\frac{2 d}{3 K \varepsilon_0 A}

Or,

\;\frac{1}{C_{\;\text{eq}\;}}\;=\;\frac{5 d}{6 K \varepsilon_0 A}

∴

C_{\;\text{eq}\;}\;=\;\frac{6 K \varepsilon_0 A}{5 d}

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