Question

Four point charges \( Q \) each, are held at the four corners of a square of side \( l \). The amount of work done in bringing a charge \( Q \) from infinity to the centre of the square will be:

• A. \(\frac{Q^2}{\pi \varepsilon_0 l} \)
• B. \( \frac{\sqrt{2 Q^2}}{\pi \varepsilon_0 l}\)
• C. \(\frac{Q^2}{2 \pi \varepsilon_0 l}\)
• D. Zero

Hint:

The work done in bringing a charge to the center of the square from infinity is the total potential at the center multiplied by the charge being moved. The symmetry of the problem plays an important role in simplifying the solution.

Answer 1

The Correct Option is B

The work done \( W \) in bringing a charge \( Q \) from infinity to the center of the square is given by the potential energy of the system. For each pair of charges, the potential at the center of the square due to a single charge \( Q \) at distance \( r \) from the center is: \[ V = \frac{Q}{4 \pi \varepsilon_0 r} \] For the four charges arranged at the corners of the square, the distance from each corner to the center is \( r = \frac{l}{\sqrt{2}} \). Thus, the potential at the center due to one charge is: \[ V = \frac{Q}{4 \pi \varepsilon_0 \frac{l}{\sqrt{2}}} = \frac{\sqrt{2} Q}{4 \pi \varepsilon_0 l} \] The total potential at the center due to the four charges is: \[ V_{\text{total}} = 4 \times \frac{\sqrt{2} Q}{4 \pi \varepsilon_0 l} = \frac{\sqrt{2} Q}{\pi \varepsilon_0 l} \] Now, the work done in bringing a charge \( Q \) from infinity to the center is given by: \[ W = Q \times V_{\text{total}} = Q \times \frac{\sqrt{2} Q}{\pi \varepsilon_0 l} \] \[ W = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 l} \] Thus, the work done in bringing the charge \( Q \) to the center of the square is: \[ W = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 l} \]