Question

Derive an expression for the electrical energy stored in a capacitor of capacitance C when charged to a potential difference V
If C = 2uF and V = 4V, calculate:
(i) the final energy stored in the capacitor (ii) the work done by the battery in the charging process. Hence account for any difference between your answers above.

Answer 1

a) Let the capacitance of the capacitor be C and it is charged to potential difference V. Let's write the potential difference:

$$Q=CV,$$ $$V=\dfrac{Q}{C}.$$

The energy stored in the capacitor equals to the work done to move the charge into the capacitor which have the potential difference V:

$$dW=VdQ=\dfrac{Q}{C}dQ.$$

Now, we can find the work done by taking the integral:

$$W=\dfrac{1}{C}\displaystyle\intop_{0}^Q QdQ=\dfrac{1}{2}\dfrac{Q^2}{C}.$$

Therefore,

$$E=W=\dfrac{1}{2}\dfrac{Q^2}{C}=\dfrac{1}{2}\dfrac{(CV)^2}{C}=\dfrac{1}{2}CV^2.$$

(i) The final energy stored in capacitor can be found as follows:

$$E=\dfrac{1}{2}CV^2=\dfrac{1}{2}\cdot2\cdot10^{-6}\ F\cdot(4\ V)^2=1.6\cdot10^{-5}\ J.$$

(ii) Let's find the total charge of the capacitor:

$$Q=CV=2\cdot10^{-6}\ F\cdot4\ V=8\cdot10^{-6}\ C.$$

Then, we can find the work done by the battery in the charging process:

$$W=QV=8\cdot10^{-6}\ C\cdot4\ V=3.2\cdot10^{-5}\ J.$$