Question:
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 the energy stored in the capacitor be two times the initial energy stored?
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 the energy stored in the capacitor be two times the initial energy stored?
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With the battery connected, V is constant so energy \(U = \frac{1}{2}CV^2 \propto C\). The capacitor with a partly inserted slab is two capacitors in parallel.
Updated On: Nov 5, 2026
- \(\frac{l}{4}\)
- \(\frac{l}{2}\)
- \(\frac{2l}{3}\)
- \(\frac{l}{3}\)
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The Correct Option is D
Solution and Explanation
Initially \(C_0 = \frac{\epsilon_0 lw}{d}\).
Let the slab (k = 4) be inside by a length x. The capacitor is equivalent to two capacitors in parallel:
slab part: \(\frac{k\epsilon_0 xw}{d}\); empty part: \(\frac{\epsilon_0(l-x)w}{d}\).
\(C = \frac{\epsilon_0 w}{d}\left[(l-x) + 4x\right] = \frac{\epsilon_0 w}{d}(l+3x) = C_0\left(1+\frac{3x}{l}\right)\).
The battery stays connected so V is constant and \(U = \frac{1}{2}CV^2 \propto C\).
\(U = 2U_0 \Rightarrow C = 2C_0 \Rightarrow 1 + \frac{3x}{l} = 2 \Rightarrow x = \frac{l}{3}\).
Answer: (4) \(\frac{l}{3}\)
Let the slab (k = 4) be inside by a length x. The capacitor is equivalent to two capacitors in parallel:
slab part: \(\frac{k\epsilon_0 xw}{d}\); empty part: \(\frac{\epsilon_0(l-x)w}{d}\).
\(C = \frac{\epsilon_0 w}{d}\left[(l-x) + 4x\right] = \frac{\epsilon_0 w}{d}(l+3x) = C_0\left(1+\frac{3x}{l}\right)\).
The battery stays connected so V is constant and \(U = \frac{1}{2}CV^2 \propto C\).
\(U = 2U_0 \Rightarrow C = 2C_0 \Rightarrow 1 + \frac{3x}{l} = 2 \Rightarrow x = \frac{l}{3}\).
Answer: (4) \(\frac{l}{3}\)
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