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 \( W \) required to move a charge \( Q \) from infinity to the square's center is equivalent to the system's potential energy. The potential \( V \) at the square's center due to a single charge \( Q \) located at a distance \( r \) from the center is calculated as: \[ V = \frac{Q}{4 \pi \varepsilon_0 r} \] For four charges at the square's corners, the distance \( r \) from each corner to the center is \( r = \frac{l}{\sqrt{2}} \). Therefore, 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 from all four charges is: \[ V_{\text{total}} = 4 \times \frac{\sqrt{2} Q}{4 \pi \varepsilon_0 l} = \frac{\sqrt{2} Q}{\pi \varepsilon_0 l} \] The work done in bringing charge \( Q \) from infinity to the center is: \[ W = Q \times V_{\text{total}} = Q \times \frac{\sqrt{2} Q}{\pi \varepsilon_0 l} \] \[ W = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 l} \] Consequently, the work done to bring the charge \( Q \) to the center of the square is: \[ W = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 l} \]