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To change the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 3/4d, where d is the separation between the plates. The new capacitance C' in terms of the original capacitance C₀ is given by

Asked in JEE Main 16th March 1st Shift 2021 · Dielectric slab

• (1) C'=(3+K)/(4K)C₀
• (2) C'=4/(3+K)C₀
• (3) C'=(4+K)/3C₀
• (4) C'=(4K)/(K+3)C₀

Step-by-step solution

Why the other options are wrong

• (1) The reciprocal of the correct factor; it would make the capacitance FALL when a dielectric is inserted, which never happens.
• (2) Inverts (K+3)/(4K) carelessly and keeps only the 4, dropping the K from the numerator, so the capacitance comes out below C₀ for every K>1, which insertion of a dielectric can never do.
• (3) Not reachable by one slip: adding the two layers in parallel gives (4(K+3))/3C₀, four times this option, and at K=1 this option gives 5/3C₀ instead of the required C₀, so it is a printed filler.

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