rge being moved, i.e.,

$ Work = q \cdot \Delta V $

where $ q $ is the charge being moved, and $ \Delta V $ is the potential difference between the initial and final points.

For a charge $ Q $ placed at a point, the potential $ V $ at a distance $ r $ from the charge is given by Coulomb's law as:

$ V = \frac{kQ}{r} $

where $ k $ is Coulomb's constant ($ k = 9 \times 10^{9} \, \mathrm{Nm^2/C^2} $).

In the case of the square PQRS with a charge of $ 100 \mu C $ at the center and sides of $1 \mathrm{~m}$, the potential at any corner due to the central charge would be the same because all corners are equidistant from the center.

Since $ PQRS $ is a square of side $1 \mathrm{~m}$, the distance from the center to any corner (the diagonal of the square, which is half the diagonal of the square) is:

$ d = \frac{\sqrt{2}}{2} $

The potential at any corner (or the work per unit charge to move a charge from infinity to that point) due to the central charge $ Q = 100 \mu C = 100 \times 10^{-6} C $ is:

$ V = \frac{kQ}{d} = \frac{9 \times 10^{9} \cdot 100 \times 10^{-6}}{\sqrt{2}/2} = 9 \sqrt{2} \times 10^{5} \, V $

Since the potential everywhere along the path (any path on the square plane) from P to R due to the central charge is constant because the distance from each point on the path to the charge is constant, the potential difference $ \Delta V $ between corners P and R is zero. Therefore, the work done in moving any charge from P to R is:

$ Work = q \cdot \Delta V = q \cdot 0 = 0 $

The correct answer is therefore:

Option C: Zero.

Electrostatics

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