Question

Case Study

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.

OR

(iii) Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in figure. Find an expression for capacitance of the system.

Answer 1

(i)

In the above circuit diagram, we can remove capacitor with plate number 33′ by applying Wheatstone bridge principle.

 

\(C_{eq} = \frac{C}{2}+ \frac{C}{2}+ C\)

\(C_{eq} = 2C\)

(ii)

 

When a dielectric slab is inserted between the charged plates of a capacitor, due to polarization of molecules of dielectric slab an internal electric field is generated in slab in the direction opposite to that of between the plates of capacitor. Hence, net electric field between the plates of capacitor decreases.

E_net = E – E_in

(iii) (a) Given C_A = C

Charge on C_A = Q

Q = CV

When two capacitors are connected in parallel then

C_eq = C + 2C

= 3C

Q_net = Q (Because charge will be conserved)

\(V' = \frac{Q}{3C}\) 

Potential difference across each capacitor \(= \frac{Q}{3C}\)

(b) Now final charge on capacitor

\(A = C \times \frac{Q}{3C}\) 

\(= \frac{QC}{3C} = \frac{Q}{3}\) 

Therefore, charge lost by capacitor \(A= Q - \frac{Q}{3} = \frac{2Q}{3}\) 

(iii)

\(C = \frac{K \varepsilon_0A}{d}\) 

\(C_1 = \frac{2K \varepsilon _0 A}{d/3} = \frac{6K\varepsilon _0 A}{d}\) 

\(C_2 = \frac{K \varepsilon _0 A}{2d/3} = \frac{3K \varepsilon _0 A}{2d}\) 

Since the two capacitors are in series therefore

\(\frac{1}{C_{eq}} = \frac{1}{C_1}+ \frac{1}{C_2}\) 

\(\frac{1}{C _{eq}} = \frac{d}{6K \varepsilon _0 A} + \frac{2d}{3K \varepsilon_0A}\)

\(\frac{1}{C_{eq}} = \frac{d}{6K \varepsilon _0 A} + \frac{4d}{6 k \varepsilon _0 A}\) 

\(\frac{1}{C_{eq} } = \frac{5d}{6K \varepsilon _0 A}\) 

\(C_{eq} = \frac{6 K \varepsilon _0 A}{5d}\)