Question

The capacitance of a charged condenser is C and energy stored on account of charge on it is U, then the quantity of charge on the condenser will be:
A) $\sqrt {2UC}$
B) $\sqrt {\dfrac{{UC}}{2}}$
C) $2UC$
D) Zero

Answer 1

When capacitor gets charged energy stored on it is given by:
$E = \dfrac{1}{2}C{V^2}$ (C is the capacitance of the capacitor and V is the voltage)
Or it can be written as:
$E = \dfrac{1}{2}\dfrac{{{Q^2}}}{C}$ (As Q=CV)
Energy E is calculated in joules, when C is calculated in farads, V is in volts and Q is in coulombs.
Using the above relations we will find the charge of the capacitor.

Complete step by step solution:
Before doing the calculation part let’s define the capacitor and its properties.
Capacitor is a passive device (which does not generate any energy) that stores energy in the form of an electric field in it. Effect shown by the capacitor is known as its capacitance. Capacitor was originally known as condenser. It is composed of two metal plates with air or dielectric filled in between the two metal plates. Various types of capacitors are Mica capacitors, Paper capacitors, Ceramic capacitors, Electrolytic capacitors etc. The name of the capacitor is denoted by its dielectric used for the formation of capacitor.

Now, comes the calculation of charge on the condenser.
In question energy is denoted by U, therefore we can write as:
$U = \dfrac{1}{2}C{V^2}$ or $U = \dfrac{1}{2}\dfrac{{{Q^2}}}{C}$..............(1)
$\Rightarrow U = \dfrac{1}{2}\dfrac{{{Q^2}}}{C}$
$\Rightarrow 2UC = {Q^2}$
$\therefore Q = \sqrt {2UC}$

We have calculated the charge by rearranging the terms of equation (1))

Option: (A) is the right answer.

Note: Charge stored on a capacitor cannot change abruptly; it takes some time to change from one state to another. And this behaviour of capacitors is called transient behaviour. When a capacitor has no charge on it acts as a short circuit in the circuit and when the capacitor gets charged it behaves as an open circuit.