Question:

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^4 \,V/m$, the charge density of the positive plate will be close to

Updated On: June 02, 2025

$6 \times 10^{-7} C/m^2 $
$3 \times 10^{-7} C/m^2 $
$3 \times 10^{4} \,C/m^2 $
$6 \times 10^{4} \,C/m^2 $

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The Correct Option is A

Solution and Explanation

$E =\frac{\sigma}{K \varepsilon_{0}} $
$\sigma =K \varepsilon_{0} E $
$=2.2 \times 8.85 \times 10^{-12} \times 3 \times 10^{4} \approx 6 \times 10^{-7} \,C / m ^{2}$

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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.