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A parallel plate capacitor has plates of length l, width w and plate separation 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 the plates will the energy stored in the capacitor be two times the initial energy stored?

Asked in JEE Main 5th Sept 2nd Shift 2020 · Dielectric slab

• (1) (2l)/3
• (2) l/3
• (3) l/4
• (4) l/2

Step-by-step solution

Why the other options are wrong

• (1) Solves x+4(l-x)=2l, i.e. treats x as the length still filled with air rather than the length covered by the slab.
• (3) Comes from l+Kx=2l: the slab is added as Kx without removing the air strip of length x that it displaces.
• (4) Guesses that the energy doubles when half the plate is covered; the capacitance is linear in x but with slope (K-1), so half coverage already gives 2.5C₀.

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