Problem 42 - A Complete Resource Book in Physics for JEE Main

Question

Four equal charges \(Q\) are placed at the four corners of a square of side \(a\). The work done in removing a charge \(-Q\) from the centre of the square to infinity is (A) Zero (B) \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\) (C) \(\frac{\sqrt{2} Q^{2}}{\pi \varepsilon_{0} a}\) (D) \(\frac{Q^{2}}{2 \pi \varepsilon_{0} a}\)

Step-by-Step Solution

Answer

The work done in removing a charge -Q from the center of the square to infinity is (B) \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\).

Step 1: 1. Find the distance between the charges

First, we need to find the distance between the charges in the center of the square and the charges at the corners. Since the charges are at the corners of a square, the diagonal of the square will be the line connecting the two opposite charges at the corners. We can use the Pythagorean theorem to determine the length of the diagonal and then find the distance between the charges. Length of the diagonal: \(d = \sqrt{2a^2}\) The distance between the charges, \(r = \frac{d}{2} = \frac{\sqrt{2a^2}}{2} = \frac{a\sqrt{2}}{2}\).

Step 2: 2. Find the potential energy of the system with the negative charge at the center

To find the potential energy, we need to multiply the Coulomb potential between the center charge and each corner charge by the magnitude of the center charge. Since we have four corner charges, we add up these potentials and then multiply by -Q. Potential energy at the center: \(U_1 = -Q \sum_{i=1}^{4} \frac{k Q}{(r_i)}\) Since all the distances \(r_i\) are equal, we can simplify this to: \(U_1= -4Q \cdot \frac{kQ}{a\sqrt{2}/2}\)

Step 3: 3. Find the potential energy of the negative charge at infinity

When the negative charge is removed to infinity, the potential energy will be zero because the separation between the charges becomes infinite. Potential energy at infinity: \(U_2 = 0\)

Step 4: 4. Compute the work done in moving the negative charge to infinity

The work done (W) in removing the negative charge from the center to infinity is the difference between the two potential energies: \(W = U_2 - U_1\) Substituting values from steps 2 and 3, we get \(W = 0 - (-4Q \cdot \frac{kQ}{a\sqrt{2}/2})\) Now, we know that \(k = \frac{1}{4\pi\varepsilon_0}\), so we can substitute this into the expression for work done: \(W = 4Q \cdot \frac{\frac{1}{4\pi\varepsilon_0}Q}{a\sqrt{2}/2}\) Simplifying this expression, we get \(W = \frac{\sqrt{2} Q^2}{4 \pi \varepsilon_{0} a}\) Therefore, the correct answer is (B) \(\frac{\sqrt{2} Q^{2}}{4 \pi \varepsilon_{0} a}\).