Solution:

Ans. According to the Gauss’s law,

$1 / 2$

$\phi=\oint \vec{E} \cdot \overrightarrow{d S}=\frac{q_{\text {enclosed }}}{\varepsilon_{0}}$ $\therefore$ Forsphere $S_{1}$, flux enclosed $=\phi_{1}=\frac{2 Q}{\varepsilon_{0}} \quad 1 / 2$ nd for sphere $S_{2}$, flux enclosed $=\phi_{2}=\frac{2 Q+4 Q}{\varepsilon_{0}}{ }_{1 / 2}$

$$ \begin{aligned} \phi_{2} & =\frac{6 Q}{\varepsilon_{0}} \ \frac{\phi_{1}}{\phi_{2}} & =\frac{1}{3} \end{aligned} $$

$1 / 2$

When a medium of dielectric constant $\varepsilon_{r}$ is introduced in the space inside sphere $S_{1}$, the flux through $S_{1}$ would be $\phi_{1}{ }^{\prime}=\frac{2 Q}{\varepsilon_{r}}$.

[CBSE Marking Scheme, 2014]

AI Q. 11. A hollow cylindrical box of length $1 \mathrm{~m}$ and area of cross-section $25 \mathrm{~cm}^{2}$ is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by $\vec{E}=50 x \hat{i}$, where $E$ is in $\mathrm{NC}^{-1}$ and $x$ is in metres. Find

(i) Net flux through the cylinder.

(ii) Charge enclosed by the cylinder.

A&E [Delhi I, II, III 2013]

Ans. (i) Given : $\vec{E}=50 x \hat{i}$ and $\Delta \mathrm{S}=25 \mathrm{~cm}^{2}$ or $25 \times 10^{-4} \mathrm{~m}^{2}$

As the electric field is only along the $x$-axis, hence, flux will pass only through the cross-section of cylinder.

Magnitude of electric field at cross-section $A$,

$$ E_{A}=50 \times 1=50 \mathrm{~N} / \mathrm{C} . $$

Magnitude of electric field at cross-section $B$,

$$ E_{B}=50 \times 2=100 \mathrm{~N} / \mathrm{C} $$

The corresponding electric fluxes are

$$ \begin{aligned} \oint \phi_{\mathrm{A}} & =\vec{E} \cdot \overrightarrow{\Delta S} \ & =50 \times 25 \times 10^{-4} \times \cos 180^{\circ} \ & =-0.125 \mathrm{Nm}^{2} / \mathrm{C}^{2} \ \oint \phi_{B} & =\vec{E} \cdot \overrightarrow{\Delta S} \ & =100 \times 25 \times 10^{-4} \times \cos 0^{\circ} \ & =0.25 \mathrm{Nm}^{2} / \mathrm{C}^{2} \end{aligned} $$

So, the net flux through the cylinder,

$$ \begin{aligned} \oint \phi & =\oint \phi_{\mathrm{A}}+\oint \phi_{B} \ & =0.125+0.25 \ & =0.375 \mathrm{Nm}^{2} / \mathrm{C}^{2} \end{aligned} $$

(ii) Using the Gauss’s law,

$$ \begin{aligned} \phi & =\oint \vec{E} \cdot \overrightarrow{d s}=\frac{q}{\varepsilon_{0}} \ \Rightarrow \quad 0.375 & =\frac{q}{8.85 \times 10^{-12}} \ \Rightarrow \quad q & =8.85 \times 10^{-12} \times 0.375 \ & =3.3 \times 10^{-12} \mathrm{C} . \ & {[C B S E \text { Marking Scheme, } 2013] } \end{aligned} $$