A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) $\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]$ (B) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]$ (C) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]$ (D) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]$

Recall the capacitance formula of a parallel plate capacitor, which is given by \(C=\varepsilon_{0}\dfrac{A}{d}\), where \(\varepsilon_{0}\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the separation between the plates.

The potential energy stored in the capacitor when it is charged to potential difference \(V_{0}\) is given by the formula \(U = \dfrac{1}{2} CV_{0}^{2}\). Using the capacitance formula, the potential energy can be expressed as \(U = \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{d})V_{0}^{2}\).

The plates are pulled apart to three times the initial separation. Let the new separation be \(d'\), then \(d' = 3d\). The capacitance after the separation is increased can be given by \(C' = \varepsilon_{0} \dfrac{A}{d'} = \varepsilon_{0} \dfrac{A}{3d}\).

Since the battery is disconnected from the capacitor, the charge on the plates is conserved and remains constant. Using the formula \(Q = CV\), and with constant \(Q\), the potential at new separation \(V'\) becomes \(V' = 3V_{0}\). Now, to find the potential energy stored in the capacitor at the new separation, we use the formula \(U' = \dfrac{1}{2} C'(V')^{2}\). Substituting the values of \(C'\) and \(V'\), we get \(U' = \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{3d}) (3V_{0})^{2}\).

The work required to separate the plates is the difference between the potential energies at the new and initial separations, i.e., \(W = U' - U\). Plugging in the expressions for \(U'\) and \(U\), we get: \(W = \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{3d}) (3V_{0})^{2} - \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{d})V_{0}^{2}\) After simplifying, we get: \(W = \dfrac{1}{2} \varepsilon_{0} A(V_{0})^{2}\left(\dfrac{1}{d}-\dfrac{1}{3d}\right)\) Which simplifies further to: \(W = \dfrac{1}{2} \varepsilon_{0} A(V_{0})^{2}\left(\dfrac{2}{3d}\right)\) Comparing the expression of work with the given options, we find that the correct option is (B). Answer: The work required to separate the plates is \(\left[\left(A \varepsilon_{0} V_{0}^{2}\right) /(2 d)\right]\).