Question

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:

• A. \( \frac{Kq_0^2}{a} (4\sqrt{2} - 2) \)
• B. \( \frac{Kq_0^2}{a} (4 - \sqrt{2}) \)
• C. \( \frac{Kq_0^2}{a} (3\sqrt{2} - 2) \)
• D. \( \frac{Kq_0^2}{a} (3 - \sqrt{2}) \)

Hint:

To solve problems involving potential energy, use the formula for the potential energy between two charges and sum over all pairs of charges. Pay attention to the distances between the charges in different configurations.

Answer 1

The Correct Option is A

The potential energy of a system of point charges is calculated as:

U = ∑_i<j \( \frac{K q_i q_j}{r_{ij}} \),

where \( r_{ij} \) is the separation between charges \( i \) and \( j \).

Configuration (1): Charges are positioned at the vertices of a square. The distance between adjacent charges is \( a \), and the distance between diagonally opposite charges is \( \sqrt{2}a \).

Configuration (2): Charges are positioned at the midpoints of the sides of the square. Consequently, the distance between adjacent charges is \( \frac{a}{\sqrt{2}} \), and the distance between diagonally opposite charges is \( a \).

The potential energy for configuration (1) is:

\( U_1 = 4 \times \frac{K q_0^2}{a} + 2 \times \frac{K q_0^2}{\sqrt{2}a} \).

The potential energy for configuration (2) is:

\( U_2 = 4 \times \frac{K q_0^2}{\frac{a}{\sqrt{2}}} + 2 \times \frac{K q_0^2}{a} \).

The difference in potential energy, calculated as \( U_2 - U_1 \), yields the required result:

\( \Delta U = U_2 - U_1 = \frac{K q_0^2}{a} (4\sqrt{2} - 2) \).

Final Answer: \( \frac{K q_0^2}{a} (4\sqrt{2} - 2) \).