Question

Derive the expression for energy stored in a charged capacitor.

Answer 1

Hint: The relation between Capacitance, charge and electrical potential. The work done to accumulate charge in a capacitor is the energy stored in a capacitor. Relation between electrical potential and work done.

Formula used: $V={\displaystyle \frac{Q}{C}}$, where, V represents the electrical potential, C represents the capacitance and Q represents the charge stored in a capacitor.
The work done to move a test charge $dQ$ is given by the expression $dW=VdQ$
Also, ${\int }_a^b{x}^{n}dx={\displaystyle \frac{b^{n+1}-a^{n+1}}{n+1}}$

Complete step by step answer:

A capacitive circuit is shown in the above figure. In a circuit, with voltage around the capacitor V, the Capacitance C is given by the equation:
$Q=CV$…(1)
Where, Q represents the charge stored in the capacitor.
Now, we want to find the energy stored in a capacitor.
According to electrostatics, the energy stored in a capacitor will be equal to the work done to move the charge into the capacitor having an electrical potential V.
Or
$dW=VdQ$…(2)
Now, for a capacitor, $V={\displaystyle \frac{Q}{C}}$…(3)
So, we can put the value of V from equation (3) into equation (2).
This gives,
$dW={\displaystyle \frac{Q}{C}}dQ$,
Now the total work done to move charge Q can be found by integration,
${\int }_0^{W}dW={\displaystyle \frac{1}{C}}{\int }_0^{Q}QdQ$
Now, from the integration method given in the formula used section, we can calculate the following result.
$$W={\displaystyle \frac{1}{2}}{\displaystyle \frac{Q^2}{C}}$$, This is also equal to the energy stored in the capacitor.
Therefore, $U={\displaystyle \frac{1}{2}}{\displaystyle \frac{Q^2}{C}}$
Moreover, from equation (1), we can put $Q^2={\left(CV\right)}^2$
This will give us,
$U={\displaystyle \frac{1}{2}}{\displaystyle \frac{Q^2}{C}}={\displaystyle \frac{1}{2}}C{V}^2$

Additional Information: A capacitor is a device in which equal and opposite charges are separated by a distance in space.
The capacitor shown in the books is usually a parallel plate capacitor.
Cylindrical capacitors are used in many electrical devices.
The SI unit is farad. It is defined as the capacitance which stores 1 C of charge along 1 V potential difference,

Note: The value of capacitance in a parallel plate is given by $C=\epsilon {\displaystyle \frac{A}{d}}$, where, $\epsilon$is the permittivity of medium between the plates, A is the area of the plates and d is the distance between the plates. The question can also be solved by substituting the value of $C=\epsilon {\displaystyle \frac{A}{d}}$, in equation (3). This will give an answer in terms of area and distance of the plates.