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Question

The net fluw linked with closed surfaces $S_{1}$, $\mathrm{S}_{2}, \mathrm{S}_{3} \& \mathrm{S}_{4}$ are

For surface $S_{1}, \phi_{1}=\frac{1}{

Vedclass · Accepted Answer

${\phi _1} = {\phi _2} = {\phi _3} = {\phi _4}$

Vedclass · Answer

The net fluw linked with closed surfaces $S_{1}$, $\mathrm{S}_{2}, \mathrm{S}_{3} \& \mathrm{S}_{4}$ are

For surface $S_{1}, \phi_{1}=\frac{1}{\varepsilon_{0}}(2 q)$

For surface

$\mathrm{S}_{2}, \phi_{2}=\frac{1}{\varepsilon_{0}}(\mathrm{q}+\mathrm{q}+\mathrm{q}-\mathrm{q})=\frac{1}{\varepsilon_{0}} 2 \mathrm{q}$

For surface $S_{3}, \phi_{3}=\frac{1}{\varepsilon_{0}}(q+q)=\frac{1}{\varepsilon_{0}}(2 q)$

For surface

$S_{4}, \phi_{4}=\frac{1}{\varepsilon_{0}}(8 q-2 q-4 q)=\frac{1}{\varepsilon_{0}}(2 q)$

Hence, $\phi_{1}=\phi_{2}=\phi_{3}=\phi_{4}$ i.e. netelectric flux is same for all surfaces.

Keep in mind, the electric field due to a charge outside $\left(\mathrm{S}_{3} 	ext { and } \mathrm{S}_{4}ight),$ the Gaussian surface contributes zero net flux through the surface, because as many lines due to that charge enter the surface as leave it.

Four closed surfaces and corresponding charge distributions are shown below

Let the respective electric fluxes through the surfaces be ${\phi _1},{\phi _2},{\phi _3}$ and ${\phi _4}$ . Then

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Four closed surfaces and corresponding charge distributions are shown below Let the respective electric fluxes through the surfaces be ${\phi _1},{\phi _2},{\phi _3}$ and ${\phi _4}$ . Then