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

Hint: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 capacitance of a conductor is$C$, and the potential difference between its plates is $V$, when connected to a battery, Energy stored in a capacitor is given by
$E = \dfrac{{C{V^2}}}{2}$
We have, the capacitance of a capacitor, $C = 20mF = 20 \times {10^{ - 3}}F$
Voltage, $V = 1000V$
Now, Energy, $E = \dfrac{{C{V^2}}}{2}$
By substituting the given values on the above equation, we get
$E = 20 \times {10^{ - 3}} \times \dfrac{{{{1000}^2}}}{2}$
We can simplify this as
$E = 10 \times {10^{ - 3}} \times {({10^3})^2}$
$ \Rightarrow E = 10 \times {10^{ - 3}} \times {10^6}$ $\left( {\because {{({a^n})}^m} = {a^{n \times m}}} \right)$
$ \Rightarrow E = 10 \times {10^3}$ $\left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right)$
$ \Rightarrow E = {10^4}J$
The potential energy of the conductor will be $ \Rightarrow E = {10^4}J$
Therefore, the correct answer is (B) ${10^4}J$
Additional information:
By using the relations, $V = \dfrac{{{Q^2}}}{{2C}} = \dfrac{1}{2}C{V^2} = \dfrac{1}{2}QV$, between Charge, Potential difference and Capacitance, we can find the equation for potential energy of a conductor in terms of all of them.
Some other useful formulae are:
$C = \dfrac{{{ \in _0}A}}{d}\& V = Ed$
$U = \dfrac{{{Q^2}}}{{2C}} = \dfrac{1}{2}\dfrac{{{ \in _0}A}}{d}{(Ed)^2}$
$U = \left( {\dfrac{1}{2}{ \in _0}{E^2}} \right)A.d$

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.