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 slab 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\(?\)

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Answer & Solution

Correct Answer

(C) \(\frac{l}{3}\)

Detailed explanation

Before inserting slab

\(C_{i}=\frac{\varepsilon_{0} A}{d}\)

\(C_{i}=\frac{\varepsilon_{0}(w)}{d}\)

After inserting dielectric slab

\(C_{f}=C_{1}+C_{2}\)

\(C_{f}=\frac{K \varepsilon_{0} A_{1}}{d}+\frac{\varepsilon_{0} A_{2}}{d}\)

\(C_{f}=\frac{K \varepsilon_{0} W x}{d}+\frac{\varepsilon_{0} w(\ell-x)}{d}\)

\(C_{f}=2 C_{i} \Rightarrow \frac{K{\varepsilon_0} w x}{x}+\frac{\varepsilon_{0} w(\ell-x)}{d}=\frac{2 \varepsilon_{0}(w)}{d}\)

\(4 x+\ell-x=2 \ell\)

\(x=\frac{\ell}{3}\)

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