A parallel plate capacitor has plate of length '$\ell$', width '$w$' and separation of plates is '$d$'. It is connected to a battery of emf V. A dielectric slap of the same thickness 'd' and of dielectric constant $\mathrm{k}=4$ is being, inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored?

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Capacitance

A parallel plate capacitor has plate of length '

ℓ

', width '

w

' and separation of plates is '

d

'. It is connected to a battery of emf V. A dielectric slap of the same thickness 'd' and of dielectric constant

k = 4

is being, inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored?

(A) $2 ℓ /3$
(B) $ℓ /2$
(C) $ℓ /4$
(D) $ℓ /3$ ✓ Correct

Answer: (D)

Solution

With the slab pushed in a length

x

, the capacitor behaves as two capacitors in parallel: a filled part

\frac{k ϵ _{0} w x}{d} = \frac{4 ϵ _{0} w x}{d}

and an empty part

\frac{ϵ _{0} w ( ℓ - x )}{d}

. Because the battery stays connected the voltage is fixed, so the stored energy

\frac{1}{2} C V^{2}

doubles exactly when the capacitance doubles. Setting

C_{f} = 2 C_{i}

gives

\frac{4 ϵ _{0} w x}{d} + \frac{ϵ _{0} w ( ℓ - x )}{d} = 2 \frac{ϵ _{0} ℓ w}{d}

, which reduces to

4 x + ℓ - x = 2 ℓ

, so

3 x = ℓ

and the slab must enter a length

x = \frac{ℓ}{3}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Capacitance

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