Question

A capacitor consists of 2 parallel metal plates in air. The distance between the plates is 5.00mm and the capacitance is 72 x10^(-12) F. The potential difference between the plates is raised to 12.0V with a battery.
(i) Calculate the energy stored in the capacitor
(ii) The battery is then disconnected from the capacitor; the capacitor retains its charge. Calculate the energy stored in the capacitor if the distance between the plates is now increased to 10.00mm.

Answer 1

(a) The energy stored in the capacitor can be calculated as follows:

$$E=\dfrac{1}{2}CV^2,$$ $$E=\dfrac{1}{2}\cdot72\cdot10^{-12}\ F\cdot(12\ V)^2=5.18\cdot10^{-9}\ J.$$

b) Let's first calculate the charge stored on the capacitor:

$$Q=CV,$$ $$Q=72\cdot10^{-12}\ F\cdot12\ V=8.64\cdot10^{-10}\ C.$$

Let's find the new capacitance of the capacitor. The capacitance can be calculated as follows:

$$C=\dfrac{\epsilon_0A}{d}.$$

From the conditions of the question we know that the separation between the plates of the capacitor is doubled, therefore, the new capacitance will be:

$$C_{new}=\dfrac{1}{2}\cdot\dfrac{\epsilon_0A}{d}=\dfrac{1}{2}\cdot C=\dfrac{1}{2}\cdot72\cdot10^{-12}\ F=36\cdot10^{-12}\ F.$$

Finally, we can find the energy stored in the capacitor:

$$E=\dfrac{1}{2}\dfrac{Q^2}{C}=\dfrac{1}{2}\cdot\dfrac{(8.64\cdot10^{-10}\ C)^2}{36\cdot10^{-12}\ F}=1.04\cdot10^{-8}\ J.$$