Question

A point charge Q is located on the axis of a disc of radius R at a distance b from the plane of the disc (figure). Show that if one-fourth of the electric flux from the charge passes through the disc, then $R=\sqrt{3}$b.

Answer 1

The total flux produced by the charge is, according to Gauss’s law, is $\frac{Q_{enclosed}}{{ϵ}_0}$. Only one quarter of this flux passes through the disk. The flux through the disk is given by:
${\varphi }_{disk}=\int E.dA$
where the integration cover the entire area of the disk. Evaluating this integral and set it equal to $\frac{Q}{4{ϵ}_0}$ relates b to R. As shown in figure, dA is the area of annular ring with radius s and width ds. The electric field at the ring make an angle $\theta$ with the normal to to the ring, the flux through the ring is:
$d{\varphi }_{ring}=E.dA=EdAcos\theta =E\left(2\pi sds\right)cos\theta$
The magnitude of the electric field has the same value at all points on the ring:
$E_{ring}=\frac{Q}{4\pi {ϵ}_0{r}^2}=\frac{Q}{4\pi {ϵ}_0(s^2+b^2)}$
and, $cos\theta =\frac{b}{\sqrt{s^2+b^2}}$
Therefore, $d{\varphi }_{ring}=\frac{Qb}{2{ϵ}_0}\frac{s}{(s^2+b^2{)}^{3/2}}ds$
To get the flux through the entire disk we integrate $d{\varphi }_{ring}$ from s = 0 to s = R
${\varphi }_{disk}={\int }_0^{R}d{\varphi }_{ring}$
After integration, ${\varphi }_{disk}=\frac{Q}{2{ϵ}_0}[1-\frac{b}{\sqrt{R^2+b^2}}]$
Since the flux through the disk is already given by $\frac{Q}{4{ϵ}_0}$, then:
$\frac{Q}{4{ϵ}_0}=\frac{Q}{2{ϵ}_0}[1-\frac{b}{\sqrt{R^2+b^2}}]$
which will give
$4b^2=R^2+b^2$
Therefore, $R=\sqrt{3}b$