Question

A point charge of 40 stat coulomb is placed 2 cm in front of an earthed plane plate of large size. Then the force of attraction on the point charge is
(A) 100 dynes
(B) 160 dynes
(C) 1600 dynes
(D) 400 dynes

Answer 1

Hint : For the plate of large size, we need to assume an infinite plane at the location. Potential of an earthed metallic plate can be considered zero. The method of imaging can be used under these conditions.

Formula used: In this solution we will be using the following formula;
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ where $F$ is the electrostatic force between two charges, ${q_1}$ and ${q_2}$ are the charges, $r$ is the distance between them, and $k$ is a constant of proportionality given as $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$ , where ${\varepsilon _0}$ is the permittivity of free space.

Complete step by step answer
A point charge of 40 stat coulomb is placed 2 cm from an earthed metallic plate of a large size as in the question. Since, the metallic plate is grounded, we can assume an equipotential surface.
An equipotential surface is a surface where the potential is constant throughout. So the potential difference is zero.
Also, since the metallic plate is of a large size compared to a point charge, it can be assumed as an infinite plane. In general, these two conditions are sufficient to use the method of imaging charges to solve the problem. In the method of imaging charges, if a positive charge is placed in front of an infinite plane conductor with voltage equal to zero, we can assume that the force of attraction is called by an equal and opposite charge at the other side of the plane at equal distance from the plane.
This implies that an equal but negative charge is at the other side of the plate at distance 2 cm from the plate, hence the total distance between them is 4cm.
Hence, the force of attraction is
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ , where ${q_1}$ and ${q_2}$ are the charges, $r$ is the distance between them, and $k$ is a constant of proportionality.
Hence, by inserting known values
$F = \dfrac{{40 \times 40}}{{{4^2}}} = \dfrac{{1600}}{{16}} = 100 \,dynes$
Hence, the answer is A.

Note
For clarity, when the unit of charge is given in stat coulomb and force in dyne, then the distance is in centimetres, and the constant of proportionality $k$ is equal to 1. This system of units is called the esu-cgs system.