Question

Derive an expression for the energy stored in a capacitor. Show that whenever two conductors share charges by bringing them into electrical contact there is loss of energy.

Answer 1

Hint: The above problem can be solved by using the concept of the charging and discharging of the capacitor. The capacitor stores the electrical energy in the electric field. The energy stored depends on the capacitance and potential difference across the plates of the capacitors.

Complete step by step answer:
Let us consider the positively charged plate of the capacitor has the charge +q and negatively charged plate has charge-q. The potential difference across the plates is V and the capacitance of the capacitor is C.
The potential difference of the capacitor is given by the ratio of magnitude of the charge on one plate to capacitance of the capacitor. The formula for potential difference is given as,
$V = \dfrac{q}{C}$
Let us consider that in the charging of the capacitor the small charge dq moves between the plates due to potential difference. The work done to move this small charge is stored as the energy of the capacitor.
The work required to move the small charge is given as,
$dW = V \cdot dq......\left( 1 \right)$
Substitute $\dfrac{q}{C}$for V in the expression (1) to find the work.
$dW = \left( {\dfrac{q}{C}} \right)dq$
$\Rightarrow dW = \dfrac{1}{C}\left( {q \cdot dq} \right)$
Integrate the above expression to find energy stored in the capacitor.
$U = \dfrac{1}{C}\int\limits_0^q {q \cdot dq} $
$\Rightarrow U = \dfrac{1}{C}\left( {\dfrac{{{q^2}}}{2}} \right)_0^q$
$\Rightarrow U = \dfrac{1}{C}\left( {\dfrac{{{q^2}}}{2} - 0} \right)$
$\Rightarrow U = \dfrac{{{q^2}}}{{2C}}......\left( 2 \right)$
Substitute $\dfrac{q}{V}$ for C in the expression (2) to find the energy stored in terms of charge and potential difference of the capacitor.
$U = \dfrac{{{q^2}}}{{2\left( {\dfrac{q}{V}} \right)}}$
$\Rightarrow U = \dfrac{{{q^2}V}}{{2q}}$
$\Rightarrow U = \dfrac{1}{2}qV$
The above expression shows that the energy stored in the capacitor depends on the charge and potential difference of the capacitor.
Whenever the charge is shared by one conductor to another conductor then potential difference is also shared to another conductor. The charge and potential difference is reduced due to sharing with other conductors and loss of energy of the first conductor occurs.
Thus, the expression for energy stored in a capacitor is given as $\dfrac{1}{2}qV$ and loss of energy of conductor occurs due to the sharing of the charge.

Note: The capacitance of the capacitor depends on the area of the plates, distance between the plates and dielectric constant of the medium between the plates of the capacitors. If the capacitance of the capacitor is high then it can store a high amount of the electrical energy.