Question

Five charges, 'q' each are placed at the corners of a regular pentagon of side 'a' as shown in figure. First, charge from 'A' is removed with other charges intact, then charge at 'A' is replaced with an equal opposite charge. The ratio of magnitudes of electric fields at O, without charge at A and that with equal and opposite charge at A is

• A. 4 : 1
• B. 2 : 1
• C. 1 : 4
• D. 1 : 2

Hint:

When solving symmetric charge distribution problems, use symmetry to reduce the problem to simpler components. The electric field vector contributions can often be simplified using the principle of superposition.

Answer 1

The Correct Option is D

In this question, the electric field at the center of the pentagon due to the charges at each vertex is asked. The geometry of the system is symmetric, and we can use the principle of superposition to find the electric field contributions at point \( O \). 

1. Case 1: Without charge at \( A \)Without the charge at \( A \), the electric fields at \( O \) due to the charges at the other four vertices (i.e., at \( B, C, D, \) and \( E \)) will contribute to the net electric field at \( O \). The magnitude of each of these fields is denoted as \( E_q \), and due to symmetry, the horizontal components cancel out, leaving only a net vertical component. 

2. Case 2: With charge at \( A \) replaced with an equal and opposite charge. When the charge at \( A \) is replaced with an equal and opposite charge, the electric field contributions at \( O \) from all five charges need to be considered. The field due to the charge at \( A \) now has the opposite direction to that of the other charges at \( B, C, D, \) and \( E \). Due to symmetry, the electric field components due to charges \( B, C, D, \) and \( E \) add up, while the field from \( A \) adds in the opposite direction. 

3. Conclusion: The net electric field in the second case will be half of the field in the first case, so the ratio of the magnitudes of the electric fields is \( 1 : 2 \). Thus, the ratio is \( 1 : 2 \).