Question

Two non-conducting plates $A$ and $B$ of radii $2R$ and $4R$ respectively are kept at distances $x$ and $2x$ from the point charge $q$. A surface cutout of a non conducting shell $C$ is kept such that its centre coincides with the point charge. Each plate and the spherical surface carries a surface charge density $\sigma$. If ${\varphi }_1$ is flux through surface of $\left(B\right)$ due to electric field of $\left(A\right)$ and ${\varphi }_2$ be the flux through $\left(A\right)$ due to electric field of $\left(B\right)$ then:

• A. ${\varphi }_1={\varphi }_2$
• B. ${\varphi }_1>{\varphi }_2$
• C. ${\varphi }_1<{\varphi }_2$
• D. It depend and on $x$ and $R$

Answer 1

The correct option is B ${\varphi }_1>{\varphi }_2$

From the defination of flux,
$\varphi =E\cdot dA$

or, $\varphi =E\left(A\right)$

where, $A$ is the area of surface.

Now, according to question,

${\varphi }_1:$ Flux through $B$ due to $A$

So, $E$ due to $A=\frac{\sigma }{2{\epsilon }_o}$ (Non- conducting charge sheet)

$\Rightarrow {\varphi }_1=E$ (Area of $B$)

$\Rightarrow {\varphi }_1=\frac{\sigma }{{\epsilon }_o}\left(\pi 4R^2\right)=\frac{2\sigma \pi R^2}{{\epsilon }_o}$

Similarly for ${\varphi }_2$: Flux through $A$ due to $B$,

$E$ due to $B=\frac{\sigma }{2{\epsilon }_o}$

$\Rightarrow {\varphi }_2=E$ (Area of $A$)

$\Rightarrow {\varphi }_2=\frac{\sigma }{2{\epsilon }_o}\left[16R^2\pi \right]=\frac{8\sigma \pi R^2}{{\epsilon }_2}$

So, ${\varphi }_2>{\varphi }_1$