The separation between the plates of a charged parallel plate capacit…

The correct answer is Option (3) → (B) and (C) only

Take the charged capacitor as isolated, so no charge can leave its plates and the charge stays constant. The capacitance depends on the separation \( d \) as \( C = \varepsilon_0 A/d \), so it falls when \( d \) increases.

(A) charge on the capacitor plates — Incorrect. The plates are isolated, so \( Q \) is constant.

(B) potential difference across the capacitor — Correct. With \( Q \) fixed and \( C \) smaller, the potential difference rises.

    \( \displaystyle V = \frac{Q}{C} = \frac{Q\,d}{\varepsilon_0 A} \propto d \)

(C) energy of the capacitor — Correct. The stored energy rises, because work is done in pulling the plates apart.

    \( \displaystyle U = \frac{Q^{2}}{2C} = \frac{Q^{2}\,d}{2\varepsilon_0 A} \propto d \)

(D) energy density between the plates of the capacitor — Incorrect. The field depends only on the charge per unit area, which does not change, so the energy density stays the same.

    \( \displaystyle E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \;\;(\text{constant}) \qquad u = \tfrac{1}{2}\varepsilon_0 E^{2} \;\;(\text{constant}) \)

Only (B) and (C) change, which is Option (3).

Note: The question does not say whether the capacitor stays connected to the battery. The answer above takes the charged capacitor as isolated. If it stays connected to the battery, the potential difference stays constant instead, and then the charge, the energy and the energy density all change, which are (A), (C) and (D), Option (2). When both readings can be defended, go by the wording: a charged capacitor with no battery mentioned is taken as isolated, so its charge stays fixed.

    \( \displaystyle Q = CV \propto \frac{1}{d} \qquad U = \tfrac{1}{2}CV^{2} \propto \frac{1}{d} \qquad u = \tfrac{1}{2}\varepsilon_0\left(\frac{V}{d}\right)^{2} \propto \frac{1}{d^{2}} \)