Four closed surfaces and corresponding charge distributions are s… - Vidyadip

MCQ

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

A
${\phi _1} < {\phi _2} = {\phi _3} > {\phi _4}$
B
${\phi _1} > {\phi _2} > {\phi _3} > {\phi _4}$
✓
${\phi _1} = {\phi _2} = {\phi _3} = {\phi _4}$
D
${\phi _1} > {\phi _3} ; {\phi _2} < {\phi _4}$

✓

Answer

Correct option: C.

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

c
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 $\left(\mathrm{S}_{3} \text { and } \mathrm{S}_{4}\right),$ the Gaussian surface contributes zero net flux through the surface, because as many lines due to that charge enter the surface as leave it.

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