Introduction

Discover how to find the electric field at the center of a square with four charges of magnitude q — three positive and one negative — on side length a.
This key CSIR NET Life Sciences physics problem uses vector superposition for precise magnitude q / (πϵ₀a²).

Problem Setup

Four charges of magnitude q are placed at the corners of a square with side length a.
Three charges are positive (+q) and one is negative (−q).
The task is to calculate the magnitude of the electric field at the square’s center using the point charge formula:

E = q / (4πϵ₀r²)

The distance from the center to each corner is r = a / √2, so r² = a² / 2.
Thus, the magnitude of the field due to each individual charge is:

E₀ = q / (2πϵ₀a²)

Vector Calculation

Let the corners be: A(0,0) +q, B(a,0) +q, C(a,a) −q, and D(0,a) +q, with the center at (a/2, a/2).
Each field vector has magnitude E₀, and unit vectors from charge positions to the center are scaled by the charge sign
(+q: field points away, −q: field points toward).

The x-components sum to:

E₀ₓ = (E₀ / √2) × (1 − 1 + 1 + 1) = 2E₀

and the y-components similarly sum to 2E₀.

Hence, net field magnitude is:

|E| = √((2E₀)² + (2E₀)²) = 2√2 E₀ = q / (πϵ₀a²).

Option Analysis

• (a) 2q / (πϵ₀a) — Incorrect; unit mismatch (field ∝ 1/a², not 1/a); overestimates by factor 2a.
• (b) q / (πϵ₀a²) — Correct; matches vector sum result (2E₀).
• (c) q / (2πϵ₀a²) — Incorrect; underestimates the net field.
• (d) 0 — Incorrect; symmetry is broken by the one negative charge, so fields add constructively along diagonals.