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Question

The capacity of a parallel plate capacitor is 10 μF. This capacitor is divided into two equal parts and these parts are filled by media of dielectric constant 2 and 4. The new capacitance will be


A
20 μF
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B
30 μF
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C
10 μF
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D
40 μF
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Solution

The correct option is B 30 μF
Given:
k1=2; k2=4; C=10 μF

As we know that capacitance of parallel plate capacitor is

C=ϵ0Ad=10 μF .....(1)

From given diagram, we can say that the two parts can be treated as two capacitors in parallel, with separation d and plate area A2.

So, effective capacitance will be

Ceq=C1+C2=k1ϵ0(A/2)d+k2ϵ0(A/2)d

Ceq=ϵ0A(k1+k2)2d

from equation (1) and given data

Ceq=10×(2+4)2=30 μF

Hence, option (b) is correct.
Key concept: Ceq=ϵ0A(k1+k2)2d

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