Question

EF at x distance due to disc

Answer 1

The electric field at a distance $x$ from the center of a uniformly charged disc of radius $R$ and surface charge density $\sigma$ along its axis is: $E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{x}{\sqrt{x^2 + R^2}} \right)$

Answer 2

The electric field ($E$) at a distance $x$ from the center of a uniformly charged disc of radius $R$ and surface charge density $\sigma$, along its axis, is given by: $E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{x}{\sqrt{x^2 + R^2}} \right)$ If the total charge $Q$ is given, then $\sigma = \frac{Q}{\pi R^2}$, leading to: $E = \frac{Q}{2\pi \epsilon_0 R^2} \left( 1 - \frac{x}{\sqrt{x^2 + R^2}} \right)$ This is derived by integrating the electric field contributions from infinitesimal rings that make up the disc.