Question

Two charges each $+ Q$ are initially at a distance $'r'$ apart. They are now moving from each other by a small distance $'x'\left( {x \ll r} \right)$ . The work done in this is proportional to
\left( 1 \right){x^2} \\\ \left( 2 \right)x \\\ \left( 3 \right)\dfrac{1}{x} \\\ \left( 4 \right)\dfrac{1}{{{x^3}}} \\\

Answer 1

Hint : In order to solve this question, we are going to find the work done by finding the force between the charge particles and the distance $'x'\left( {x \ll r} \right)$ which is already given and then their product, which gives the value of the work done , by using the fact that $x \ll r$ , the proportionality is found out.
The force between the two charges $+ Q$ separated by a distance $'r'$ is given by
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}}$
And the work done by the particle is given by
$W = F \cdot x$

Complete Step By Step Answer:
We are given that the two charges $+ Q$ are initially, at a distance $'r'$ apart,
They move from each other by a small distance $'x'\left( {x \ll r} \right)$ ,
Work done is given by the product of force and the displacement, here the force is given by
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}}$
The displacement of the charges is $x$
Therefore, the work done is
W = F \cdot x \\\ \Rightarrow W = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}} \cdot x \\\
Now as $x \ll r$ , we can take the factor ${r^2}$ as the constant.
Therefore, the work done is directly proportional to $x$ ,
Hence, option $\left( 2 \right)x$ is the correct answer.

Note :
The work done on any particle depends upon the force acting on them and the displacement that is produced in them, here in this case, the force is constant as the charge values are constant and the initial distance by which they are apart is also constant, but the displacement $x$ produced is variable, that’s why the work done is proportional to that only.