Question

A 12µF capacitor is required to store 4J of energy. Find the p.d to which the capacitor must be charged

Answer 1

The energy $W$ , that can be stored into a capacitor is given by the expression:

$W = \dfrac{CU^2}{2}$ ,

where $C$ is the capacitance and $U$ is the potential diffectnce (p.d).

Express $U$ from this formula and substitute numerical values:

$U = \sqrt{\dfrac{2W}{C}} = \sqrt{\dfrac{2\cdot 4J}{12\cdot 10^{-6}F}} = 816.5 V$ .

Answer. 816.5 V.