Question:

In a parallel plate capacitor, if the intervening medium of permittivity $\varepsilon$ between the plates is replaced by another medium of permittivity $\frac{\varepsilon}{2} ,$ then its capacitance is

KEAM

Updated On: Jun 8, 2024

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doubled
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The Correct Option is A

Solution and Explanation

The capacitance of a parallel plate capacitor having a medium of permittivity e between the plates is
$C = \frac{\varepsilon A}{d}\,\,\,....(i) $
where $d$ is the distance between the plates and $A$ is the area of each plate.
When the medium of this capacitor is replaced by another medium of permittivity $\frac{\varepsilon}{2}$ then its capacitance becomes
$ C' = \frac{\varepsilon}{2} \frac{A}{d} $
$ = \frac{1}{2} (\frac{\varepsilon A}{d}) = \frac{C}{2}\,\,\,$ (Using $(i)$)

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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.