Question 502: Electrostatics & Capacitance

Question

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 :
![](https://imagex.cdn.examgoal.net/skJJKOAlDMFMdbJO3/bgvT95AtkxXjdH6TpTwccbRhBIyhR/oFZpiAfqgvNTMCTFO5CAyQ/image.png)

Accepted Answer

Net flux through a closed surface,
$\phi $ = ${{{q_{enclose}}} \over {{\varepsilon _0}}}$
q
enclosed
= charge enclosed by closed surface.
For surface S
1
,
$\phi $
1
= ${1 \over {{\varepsilon _0}}}$ (2q)
For surface S
2
,
$\phi $
2
=  ${1 \over {{\varepsilon _0}}}$ (q + q + q $-$ q) = ${1 \over {{\varepsilon _0}}}$(2q)
For Surface S
3
,
$\phi $
3
= ${1 \over {{\varepsilon _0}}}$ (q + q) = ${1 \over {{\varepsilon _0}}}$ (2q)
For surface S
4
,
$\phi $
4
= ${1 \over {{\varepsilon _0}}}$ (8q $-$ 2q $-$ 4q) = ${1 \over {{\varepsilon _0}}}$ (2q)
$	herefore\,\,\,$ $\phi $
1
= $\phi $
2
= $\phi $
3
= $\phi $
4

Answer

${\Phi _1}$ < ${\Phi _2}$ = ${\Phi _3}$  >  ${\Phi _4}$

Answer

${\Phi _1}$ > ${\Phi _2}$ > ${\Phi _3}$  >  ${\Phi _4}$

Answer

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

Answer

${\Phi _1}$ > ${\Phi _3}$ ; ${\Phi _2}$  <  ${\Phi _4}$

Electrostatics & Capacitance

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 :

Choose the correct answer: