Question

The surface charge density of a thin charged disc of radius R is σ. The value of the electric field at the centre of the disc is (σ)/(2varepsilon₀). With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc:

• A. reduces by 70.7%
• B. reduces by 29.3%
• C. reduces by 9.7%
• D. reduces by 14.6%

Hint:

For axial electric fields of discs, memorize: E = (σ)/(2varepsilon₀) (1 - fracx√(x² + R²)) It helps quickly compare fields at different points.

Answer 1

The Correct Option is B

Step 1: Electric field on the axis of a uniformly charged disc at distance x: E = (σ)/(2varepsilon₀) (1 - fracx√(x² + R²))
Step 2: At the centre (x=0): E₀ = (σ)/(2varepsilon₀)
Step 3: At x = R: ER = (σ)/(2varepsilon₀) (1 - \frac1√(2))
Step 4: Fractional reduction: (E₀ - ER)/(E₀) = \frac1√(2) ≈ 0.293
Step 5: Percentage reduction: 29.3%