Question

The energy of a charged conductor is given by the expression:
A. $\dfrac{{{q^2}}}{{2C}}$
B. $\dfrac{{{q^2}}}{C}$
C. 2qC
D. $\dfrac{q}{{2{C^2}}}$, where q is charge on the conductor, C is its capacitance.

Answer 1

Hint: This question is based on the concept of energy of capacitor. Capacitor is an electrical component that has the ability/capacity to store energy in the form of electric potential energy.

Complete step by step solution:
Two charged conductors (one positive and the other negative) placed close to each other constitutes a capacitor. The charge on the positive plate is called the charge on the capacitor (Q).
The potential difference between the two plates is called the potential of the capacitor (V).
For any given capacitor, charge (Q) on the capacitor is directly proportional to the potential of the capacitor (V).

$ \Rightarrow \dfrac{Q}{V} = {\text{constant}}$

This constant is called the capacitance of the capacitor, and is denoted by ‘C’.

$\dfrac{Q}{V} = C$
$ \Rightarrow Q = CV \to (1)$

The formula to determine the capacitance (For parallel plate)

$C = \dfrac{{{\varepsilon _o}A}}{d} \to (2)$

Where, A is the area of each plates, and d is the distance of separation between the two plates

Force between two plates of a capacitor is given by

$F = \dfrac{{{Q^2}}}{{2A{\varepsilon _o}}} \to (3)$

Work to be done in bringing the two plates at a distance d with respect to each other is:

$W = Fd = \left( {\dfrac{{{Q^2}}}{{2A{\varepsilon _o}}}} \right)d$

We know that, $d = \dfrac{{{\varepsilon _o}A}}{C} \to ({\text{from equation (}}2))$

$W = \left( {\dfrac{{{Q^2}}}{{2A{\varepsilon _o}}}} \right)\left( {\dfrac{{{\varepsilon _o}A}}{C}} \right)$
$W = \dfrac{{{Q^2}}}{{2C}}$

This work done is stored as energy in the capacitor.
Hence, $U = \dfrac{{{Q^2}}}{{2C}}$

Therefore, the correct choice is option (A).

Note: Using $Q = CV$ we get $U = \dfrac{{{C^2}{V^2}}}{{2C}} = \dfrac{{C{V^2}}}{2}$
Also, $CV - Q \Rightarrow U = \dfrac{{QV}}{2}$