The energy and capacity of a charged parallel plate capacitor are \(E\) and \(C\) respectively. If a dielectric slab of ${}_{}$\(E_r=6\) is inserted in it, then the energy and capacity become:
(Assuming the charge on plates remains constant)

Energy per unit volume for a capacitor having area \(A\) and separation \(d\) kept at a potential difference \(V\) is given by:
1. \(\frac{1}{2}\varepsilon_0\frac{V^2}{d^2}\)
2. \(\frac{1}{2}\frac{V^2}{\varepsilon_0d^2}\)
3. \(\frac{1}{2}CV^2\)
4. \(\frac{Q^2}{2C}\)