Question

The two circular surfaces of a frustum cut out of a right circular cone have radii $2R$ and $3R$ with their centres at $A$ and $B$ respectively and height, $h=3R$. When a point charge $Q$ is placed at $A$, the flux of electric field through the circular face of radius $3R$ is ${\varphi }_1$, and when the charge $Q$ is placed at $B$ , the flux through the circular face of radius $2R$ is ${\varphi }_2$.
The ratio $\frac{{\varphi }_1}{{\varphi }_2}$ is :

• A. $\left[\frac{\sqrt{2}-1}{\sqrt{13}-3}\right]\sqrt{\frac{2}{13}}$
• B. $\left[\frac{\sqrt{2}-1}{\sqrt{13}-3}\right]\sqrt{\frac{13}{2}}$
• C. $\left[\frac{\sqrt{13}-3}{\sqrt{2}-1}\right]\sqrt{\frac{2}{13}}$
• D. $\left[\frac{\sqrt{13}-3}{\sqrt{2}-1}\right]\sqrt{\frac{13}{2}}$

Answer 1

The correct option is B $\left[\frac{\sqrt{2}-1}{\sqrt{13}-3}\right]\sqrt{\frac{13}{2}}$


When charge is at $A$, the flux through face of radius $3R$ is,
${\varphi }_1=\frac{q}{{\in }_0}\left[\frac{2\pi (1-\cos \theta )}{4\pi }\right]=\frac{q}{2{\in }_0}[1-\cos \theta ]$
$=\frac{q}{2{\in }_0}[1-\frac{3R}{{\sqrt{{\left(3R\right)}^2+\left(3R\right)}}^2}]=\frac{q}{{\in }_0}[1-\frac{\text{/}3R}{\text{/}3R\sqrt{2}}]$
$\Rightarrow {\varphi }_1=\frac{q}{{\in }_0}[1-\frac{1}{\sqrt{2}}]$
Similarly when charge is at $B$, the flux through face of radius $2R$ is ,

${\varphi }_2=\frac{q}{2{\in }_0}[1-\frac{3R}{\sqrt{{\left(3R\right)}^2+{\left(2R\right)}^2}}]$
$\Rightarrow {\varphi }_2=\frac{q}{2{\in }_0}[1-\frac{3R}{\sqrt{13R^2}}]=\frac{q}{2{\in }_0}[1-\frac{3}{\sqrt{13}}]$
$\Rightarrow \frac{{\varphi }_1}{{\varphi }_2}=\frac{\left(\frac{q}{2{\in }_0}\right)[1-\frac{1}{\sqrt{2}}]}{\left(\frac{q}{2{\in }_0}\right)[1-\frac{3}{\sqrt{13}}]}=\frac{\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}{\left(\frac{\sqrt{13}-3}{\sqrt{13}}\right)}=\left(\frac{\sqrt{2}-1}{\sqrt{13}-3}\right)\sqrt{\frac{13}{2}}$