JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

In a parallel plate capacitor set up, the plate area of capacitor is \(2 \,m ^{2}\) and the plates are separated by \(1\, m\). If the space between the plates are filled with a dielectric material of thickness \(0.5\, m\) and area \(2\, m ^{2}\) (see \(fig.\)) the capacitance of the set-up will be \(.........\, \varepsilon_{0}\) (Dielectric constant of the material \(=3.2\) ) and (Round off to the Nearest Integer)

A \(1\)
B \(5\)
C \(3\)
D \(6\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

\(C=\frac{\varepsilon_{0} A }{\frac{ d }{2 K }+\frac{ d }{2}}=\frac{2 \varepsilon_{0} A }{\frac{ d }{ K }+ d }\) \(=\frac{2 \times 2 \varepsilon_{0}}{\frac{1}{3.2}+1}=\frac{4 \times 3.2}{4.2} \varepsilon_{0}\) \(=3.04 \,\varepsilon_{0}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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