Question

If the value of electric field just outside the surface of a uniformly charged disc of radius $4\text{m}$, is $10\text{N/C}$. Then the electric field due to the disc at point lying on the axis of the disc, at a distance of $3\text{m}$ from the center will be :

• A. $2\text{N/C}$
• B. $4\text{N/C}$
• C. $8\text{N/C}$
• D. $\text{Zero}$

Answer 1

The correct option is B $4\text{N/C}$

We know that electric field just outside a charged surface is given by:
$E_o=\frac{\sigma }{2{ϵ}_o}$

$\Rightarrow \sigma =2{ϵ}_o{E}_o$ -------$\left(1\right)$

Electric field on the axis of the disc is given by :
$E=2\pi k\sigma [1-\frac{z}{\sqrt{z^2+R^2}}]$

Here symbols have usual meaning and $z$ is the distance of the point from the center of the disc.

$\Rightarrow E=2\pi \times \frac{1}{4\pi {ϵ}_o}\times 2{ϵ}_o{E}_o[1-\frac{z}{\sqrt{z^2+R^2}}]$
$\left[\text{From}\right(1\left)\right]$

$E=E_o[1-\frac{z}{\sqrt{z^2+R^2}}]$

From the data given in the question :
$E=10[1-\frac{3}{\sqrt{3^2+4^2}}]$

$\Rightarrow E=4\text{N/C}$