Derive an expression for energy stored in a charged capacitor.A spherical metal ball of radius 15 cm carries a charge of 2µC. Calculate the electric field at a distance of 20 cm from the center of the sphere.

Answer: \(U = \dfrac{Q^2}{2C} = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV\); \(E \approx 4.5\times10^5\) N/C (V/m).

Energy stored: Consider a capacitor of capacitance \(C\) being charged. At some instant let the charge on it be \(q\), so the potential difference is \(V' = \dfrac{q}{C}\).

Work done to transfer a further small charge \(dq\) from one plate to the other: \(dW = V'\,dq = \dfrac{q}{C}\,dq\).

Total work to charge it from 0 to \(Q\): \(W = \displaystyle\int_0^Q \dfrac{q}{C}\,dq = \dfrac{1}{C}\left[\dfrac{q^2}{2}\right]_0^Q = \dfrac{Q^2}{2C}\).

This work is stored as electrostatic potential energy in the field between the plates: \(U = \dfrac{Q^2}{2C}\). Using \(Q = CV\): \(U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV\).

Numerical: \(q = 2\times10^{-6}\) C, \(R = 0.15\) m, \(r = 0.20\) m. Since \(r \gt R\), the charged sphere behaves as a point charge at its centre:

\(E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2} = 9\times10^9\times\dfrac{2\times10^{-6}}{(0.2)^2} = \dfrac{1.8\times10^4}{0.04} = 4.5\times10^5\) N/C,

directed radially outward. (Using \(\varepsilon_0 = 8.85\times10^{-12}\) and \(\pi = 3.142\) gives \(4.49\times10^5\) N/C.)