Question 619: Electrostats & 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 :

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

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

Electrostats & 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: