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.
Hence account for the difference of your answer in the above two cases.

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:

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

As we can see from the calculations, when the battery is disconnected from the capacitor and the separation between plates is doubled, capacitance is halved. As a result, the energy stored in the capacitor increases.