Question

The surface charge density of a thin charged disc of radius \( R \) is \( \sigma \). The value of the electric field at the centre of the disc is \( \dfrac{\sigma}{2\varepsilon_0} \). With respect to the field at the centre, the electric field along the axis at a distance \( R \) from the centre of the disc is:

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

Hint:

For discs, remember: centre field is maximum and decreases gradually along the axis.

Answer 1

The Correct Option is B

Step 1: Electric field on the axis of a charged disc:
\( E = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{x}{\sqrt{x^2 + R^2}}\right) \) 

Step 2: At \( x = R \):
\( E = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{1}{\sqrt{2}}\right) \) 

Step 3: Fractional reduction:
\( \dfrac{1}{\sqrt{2}} \approx 0.707 \Rightarrow \text{reduction} \approx 29.3\% \)