Question

Consider two hollow concentric spheres, $S_1$ and $S_2$, enclosing charges $2Q$ and $4Q$ respectively as shown in the figure. (i) Find out the ratio of the electric flux through them. (ii) How will the electric flux through the sphere $S_1$ change if a medium of dielectric constant ${}^{\prime }ϵr^{\prime }$ is introduced in the space inside $S_1$ in place of air? Deduce the necessary expression.

Answer 1

Given two hollow concentric spheres $S_1$ and $S_2$ enclosing charges $Q$ and $2Q$.

i) We have to find the ratio of electric flux through them.

So, from Gauss law, flux through a surface is equal to \dfrac{Total\;charge\;enclosed}{\varepsilon _{0}}

So, electric flux passing through sphere $S_1$

${\varphi }_1=\frac{2Q}{{\epsilon }_0}$

And, electric flux passing through sphere $S_2$

${\varphi }_2=\frac{2Q+4Q}{{\epsilon }_0}=\frac{6Q}{{\epsilon }_0}$

$\therefore$ Ratio of flux through $S_1$ and $S_2$ is

$\frac{{\varphi }_1}{{\varphi }_2}=\frac{2Q\times {\epsilon }_0}{{\epsilon }_0\times 6Q}$

$\frac{{\varphi }_1}{{\varphi }_2}=\frac{1}{3}$

ii) Now, we have to find the electric flux through sphere $S_1$ if a medium of dielectric constant '${\epsilon }_r$' is introduced in space inside $S_1$ in place of air.

Using Gauss theorem,

${\varphi }_1=\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{2Q}{{\epsilon }_0}$

Now when a material with dielectric constant ${\epsilon }_r$ is introduced then, ${\epsilon }_r=\frac{\epsilon }{{\epsilon }_0}$

Now, flux through $S_1$ is ${\varphi }_1^{\prime }=\frac{2Q}{\epsilon }$

But $\epsilon ={\epsilon }_r{\epsilon }_0$

Where $\epsilon =$ permittivity of medium

${\epsilon }_0=$ permittivity of air

So, ${\varphi }_1^{\prime }=\frac{2Q}{{\epsilon }_r{\epsilon }_0}$

But $\frac{2Q}{{\epsilon }_0}={\varphi }_1$

So, ${\varphi }_1^{\prime }=\frac{{\varphi }_1}{{\epsilon }_r}$

So, now flux is reduced ${\epsilon }_r$ times when placed in a dielectric medium of dielectric constant ${\epsilon }_r$