Question

A capacitor of capacitance ${\text{C}}$ has a charge of ${\text{Q}}$ stored on the plates. The potential difference between the plates is doubled. What is the change in the energy stored by the capacitor?

Answer 1

Hint: A capacitor is a device that is used to store charges in an electrical circuit. A capacitor works on the principle that the capacitance of a conductor increases appreciably when an earthed conductor is brought near it. A capacitor has two parallel plates having equal and opposite charges on its surface.


Complete step by step answer:
Given,
Capacitance $ = {\text{C}}$
Charge $ = {\text{Q}}$
Let energy be ${\text{E}}$ and voltage across the capacitor ${\text{V}}$ .
So, now ${\text{E = }}\dfrac{1}{2}\dfrac{{{{\text{Q}}^2}}}{{\text{C}}}\, = \,\dfrac{1}{2}{\text{QV = }}\dfrac{1}{2}{\text{C}}{{\text{V}}^2}$
Now if the potential difference between the plates is doubled, ${{\text{V}}_1}{\text{ = 2V}}$
Therefore, the energy becomes, $${{\text{E}}_1}{\text{ = }}\dfrac{1}{2}{\text{C}}\,{{\text{(2V)}}^2}$$
Now to find the energy difference or change in the energy stored by the capacitor we have to find the difference between the previous energy and the energy stored after changing potential difference.
$ \Rightarrow \,\Delta {\text{E = }}{{\text{E}}_1} - {\text{E}}$
$ \Rightarrow \,\Delta {\text{E = }}\dfrac{1}{2}{\text{C (2V}}{{\text{)}}^2} - \,\,\dfrac{1}{2}{\text{C}}{{\text{V}}^2}$
$ \Rightarrow \,\Delta {\text{E = }}\dfrac{1}{2}{\text{C }}{{\text{V}}^2}(4 - 1)$
$ \Rightarrow \,\Delta {\text{E = }}\dfrac{3}{2}{\text{C }}{{\text{V}}^2}$
Therefore, the change in energy is $\Delta {\text{E = }}\dfrac{3}{2}{\text{C }}{{\text{V}}^2}$.

Note: Capacitor is an energy storing device. Note that capacitance of the capacitor does not depend on the potential difference across the capacitor. As the capacitance is the ratio of the charge and voltage applied, the ratio remains constant for the particular parallel plate capacitor. Hence, we also use this property in this question.