Question

A conductor of capacity 20mF is charged to 1000 V. The potential energy of the conductor will be:
(A) $20 \times {10^4}J$
(B) ${10^4}J$
(C) $20 \times {10^3}J$
(D) ${10^3}J$

Answer 1

A capacitor is a device for storing energy. When we connect a battery across the two plates of a capacitor, the current charges the capacitor, leading to an accumulation of charges on opposite plates of a capacitor.

Complete step by step answer:
If the capacitor of the conductor is$C$, it is uncharged initially and the potential difference between its plates is $V$, when connected to a battery.
Energy stored in a capacitor$= E = \dfrac{{C{V^2}}}{2}$
Given, Capacitor $= 20mF = 20 \times {10^3}$
Voltage V$= 1000V$
Here,
Energy$= \dfrac{{C{V^2}}}{2}$
$= 20 \times {10^{ - 3}} \times \dfrac{{{{1000}^2}}}{2}$
$= {10^4}J$
$= 10kJ$
Hence, the energy stored in the conductor will be 10 kilo joules.
Energy of a charged conductor or capacitor-
$\to$ If C is the capacitance of a conductor or capacitor & V is the potential of the conductor (P.D in the can of a capacitor). Then the stored electrostatic energy is:
$V = \dfrac{{{Q^2}}}{{2C}} = \dfrac{1}{2}C{V^2} = \dfrac{1}{2}QV$
$C = \dfrac{{{ \in _0}A}}{d}\& V = Ed$
$U = \dfrac{{{Q^2}}}{{2C}} = \dfrac{1}{2}\dfrac{{{ \in _0}A}}{d}{(Ed)^2}$
$U = \left( {\dfrac{1}{2}{ \in _0}{E^2}} \right)A.d$
Energy during = Energy permit volumes
$u = \dfrac{U}{{volume}}$
$u = \dfrac{1}{2}{ \in _0}{E^2}$
If dielectric is introduced.
$u = \dfrac{1}{2}k{ \in _0}{E^2}$
This energy is stored in a capacitor in the electric field below its plates.

So, the correct answer is “Option B”.

Note:
A capacitor is also called a condenser. A capacitor incorporated in an alternating current (A.C) Circuit is alternately charged and discharged thus depends on the frequency of current and if the time required is greater than the length of the half cycle the polarization i.e. separation of charge is not complete. Under such conditions the dielectric constant appears to be less than that observed in a direct current circuit and to vary with frequency becoming lower at higher frequency.