Question

A disc of radius $R$ is uniformly charged with total charge $Q$. Find the direction and magnitude of the electric field at the point $P$ lying a distance $x$ from the centre of the disc along the axis of symmetry of the disc. Given- $k=\frac{1}{4\pi {ϵ}_0}\sigma =\frac{Q}{\pi R^2}$

• A. $2\pi k\sigma [1-\frac{x}{(x^2+R^2{)}^{3/2}}]$
• B. $\frac{\sigma }{2{ϵ}_0}[1-\frac{x}{(x^2+R^2{)}^{3/2}}]$
• C. $\frac{Q}{2\pi {ϵ}_0{R}^2}[1-\frac{x}{(x^2+R^2{)}^{3/2}}]$
• D. All of the above

Answer 1

The correct option is D All of the above

Let the disk be broken into the rings of radius $R$, and thickness $dr$.

Total charge of a ring is $\sigma 2\pi \sigma .dr$$⟹\lambda 2\pi r$

$\therefore \lambda$ is the line charge of each ring.

$E_{Ring}=\frac{1}{4\pi E_c}\frac{\left(\sigma dr\right)2\pi rx}{(r^2+x^2{)}^{3/2}}$

$\overrightarrow{E_{disk}}=\frac{1}{4\pi {ϵ}_0}2\pi \sigma x{\int }_R^0\frac{r}{(r^2+x^2{)}^{3/2}}dr$

$⟹\frac{1}{4\pi {ϵ}_0}2\pi \sigma x[\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}]\hat{x}$

$⟹\frac{2\pi \sigma }{4\pi {ϵ}_0}[1-\frac{x}{\sqrt{R^2+x^2}}]\hat{x}$

$⟹2\pi k\sigma [1-\frac{x}{{(x^2+R^2)}^{3/2}}]$

or, $\frac{\sigma }{2{ϵ}_0}[1-\frac{x}{{(x^2+R^2)}^{3/2}}],$ or, $\frac{Q}{2\pi {ϵ}_0{R}^2}[1-\frac{x}{{(x^2+R^2)}^{3/2}}]$