JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

A parallel-plate capacitor has plate area

20 \mathrm{~cm}^2

, plate separation

1.0 \mathrm{~mm}

and a dielectric slab of dielectric constant

5.0

filling up the space between the plates. This capacitor is joined to a battery of emf

6.0 \mathrm{~V}

through a

100 \mathrm{k} \Omega

resistor. Find the energy of the capacitor

8.9 \mu \mathrm{s}

after the connections are made.

View answer and explanation

Correct option: A

\frac{530.97 \times 530.97}{88.5 \times 2} \times 10^{12}

Option A (

\frac{530.97 \times 530.97}{88.5 \times 2} \times 10^{12}

) is correct because it is the energy stored in the capacitor calculated using

U = \frac{1}{2}CV^2

after considering the charging process through the resistor.

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\mathrm{\rho }

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