The energy stored in a given capacitor is given by E = \(\frac{1}{2}\) CV\(^2\) = \(\frac{1}{2}\) QV = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\) Where C = the capacitance of the capacitor Q = charge carried by the capacitor V = p.d across the plates of the capacitor, thus from the data given, C = 10\muF = 10 x 10\(^{-6}\)F Q = 100\muc = 100 x 10\(^{-6}\)C ∴ Energy stored = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\) = \(\frac{1}{2}\)\(\frac{(100 \times 10^{-6})^2}{10 \times 10^{-6}}\) = 0.0005 = 5 x 10\(^{-4}\)J