Four equal charges \(\mathrm{Q}\) are placed at the four corners of a square of each side is ' \(\mathrm{a}\) '. Work done in removing a charge \- Q from its centre to infinity is ....... (A) 0 (B) $\left[\left(\sqrt{2} \mathrm{Q}^{2}\right) /\left(\pi \epsilon_{0} \mathrm{a}\right)\right]$ (C) $\left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]$ (D) \(\left[\mathrm{Q}^{2} /\left(2 \pi \epsilon_{0} \mathrm{a}\right)\right]\)

To calculate the electric potential at the center of the square, first, we need to find the distance between the center and one of the corners. Recalling that the diagonal of a square can be found using the Pythagorean theorem, we have: \(d = \sqrt{a^2 + a^2}\) As all sides of the square are equal in length, we have: \(d = \sqrt{2a^2} = a\sqrt{2}\) Now, we can find the electric potential at the center due to one of the corner charges using the formula: \(V = \frac{kQ}{r}\) Where \(k = \frac{1}{4πε_0}\) and \(r = a\sqrt{2}\). We have four equal charges Q at the corners, so the total potential at the center (V_total) is: \(V_{total} = 4 \times \frac{kQ}{a\sqrt{2}}\)

The work done to move a charge against an electric field is given by the product of the charge and the electric potential difference. In this case, the potential difference is given by (V_total - 0) since the potential at infinity is 0. Therefore, the work done (W) is given by: \(W = (-Q) \times \left(4 \times \frac{kQ}{a\sqrt{2}} - 0\right)\) Simplify the expression: \(W = \left(-\frac{4kQ^2}{a\sqrt{2}}\right)\) Now, substituting the value of \(k = \frac{1}{4πε_0}\): \(W = \left(\frac{-4Q^2}{4πε_0a\sqrt{2}}\right)\) Thus, the work done in removing a charge -Q from the center of the square to infinity is: \(W = \left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]\) So, the correct answer is (C).