Question:

Two plates of a parallel plate capacitor of capacity $ 50\mu F $ are charged by a battery to a potential of 100 V. The battery remains connected and the plates are separated from each other so that the distance between them is doubled. The energy spent by the battery in doing so, will be

Updated On: Jun 23, 2023

$ 12.5\times {{10}^{-2}}J $
$ -25\times {{10}^{-2}}J $
$ 25\times {{10}^{-2}}J $
$ -12.5\times {{10}^{-1}}J $

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The Correct Option is A

Solution and Explanation

We know that when separation between the plates is doubled the capacitance becomes one half i.e., $ C=25\,\mu F $ The energy spend by the battery is given by $ =qV=(CV)V=C{{V}^{2}} $ $ =25\times 10-6\times {{(100)}^{2}} $ $ =25\times {{10}^{-2}} $

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Notes on electrostatic potential and capacitance

Concepts Used:

Electrostatic Potential and Capacitance

Electrostatic Potential

The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.

Some major things that we should know about electric potential:

They are denoted by V and are a scalar quantity.
It is measured in volts.

Capacitance

The ability of a capacitor of holding the energy in form of an electric charge is defined as capacitance. Similarly, we can also say that capacitance is the storing ability of capacitors, and the unit in which they are measured is “farads”.

The capacitor is in Series and in Parallel as defined below;

In Series

Both the Capacitors C₁ and C₂ can easily get connected in series. When the capacitors are connected in series then the total capacitance that is C_totalis less than any one of the capacitor’s capacitance.

In Parallel

Both Capacitor C₁ and C₂ are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is C_total is any one of the capacitor’s capacitance.