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

A parallel plate capacitor of capacitance \(1 \mu \mathrm{~F}\) is charged to a potential difference of 20 V . The distance between plates is \(1 \mu \mathrm{~m}\). The energy density between plates of capacitor is _________.

A \(2 \times 10^{-4} \mathrm{~J} / \mathrm{m}^3\)
B \(1.8 \times 10^5 \mathrm{~J} / \mathrm{m}^3\)
C \(1.8 \times 10^3 \mathrm{~J} / \mathrm{m}^3\)
D \(2 \times 10^2 \mathrm{~J} / \mathrm{m}^3\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(1.8 \times 10^3 \mathrm{~J} / \mathrm{m}^3\)

Step-by-step Solution

Detailed explanation

\begin{aligned} \text { Energy density } & =\frac{1}{2} \varepsilon_0 E^2 \\ & =\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2 \\ & =\frac{1}{2}\left(8.85 \times 10^{-12}\right)\left(\frac{20}{10^{-6}}\right)^2 \mathrm{~J} / \mathrm{m}^3 \\ & \simeq 1.8 \times 10^3…

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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