Question

A point charge placed on the axis of a uniformly charged disc experiences a force F due to the disc. If the surface charge density on the disc is $\sigma$, the electric flux through the disc due to the point charge will be :

• A. $\frac{2\pi F}{\sigma }$
• B. $\frac{F}{2\pi \sigma }$
• C. $\frac{2F}{\sigma }$
• D. $\frac{F}{\sigma }$

Answer 1

Answer: (D)

$\frac{F}{\sigma }$

The electric field due to the charged disc for an axial point, $E_d=\frac{\sigma }{2{ϵ}_0}(1-\frac{x}{\sqrt{x^2+R^2}})=\frac{\sigma }{2{ϵ}_0}(1-\cos {\theta }_0)$

Force on q, $F=q{E}_d=\frac{q\sigma }{2{ϵ}_0}(1-\cos {\theta }_0).....\left(1\right)$

Consider a ring of radius $y$ and thickness $dy$.

The flux through this ring,$d\varphi =\left(E\cos \theta \right)2\pi ydy$ (as only horizontal component contribute flux and flux due to vertical component is zero because $E_y\hat{j}.\hat{i}=0$ )

Total flux,$\varphi ={\int }_0^R\left(\frac{q}{4\pi {ϵ}_0{r}^2}\cos \theta \right)2\pi ydy$

$\Rightarrow \varphi =\frac{q}{2{ϵ}_0}{\int }_0^R\frac{1}{x^2+y^2}\frac{x}{\sqrt{x^2+y^2}}ydy$

$\varphi =\frac{qx}{2{ϵ}_0}{\int }_0^R\frac{y}{(x^2+y^2{)}^{3/2}}dy=\frac{q}{2{ϵ}_0}[1-\frac{x}{\sqrt{x^2+R^2}}]$

$\Rightarrow \varphi =\frac{q}{2{ϵ}_0}(1-\cos {\theta }_0)....\left(2\right)$

From (1), (2) becomes,$\varphi =\frac{F}{\sigma }$