JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 27)
Four closed surfaces and corresponding charge distributions are shown below.
_9th_April_Morning_Slot_en_27_1.png)
Let the respective electric fluxes through the surfaces be $${\Phi _1},$$ $${\Phi _2},$$ $${\Phi _3}$$ and $${\Phi _4}$$. Then :
Let the respective electric fluxes through the surfaces be $${\Phi _1},$$ $${\Phi _2},$$ $${\Phi _3}$$ and $${\Phi _4}$$. Then :
$${\Phi _1}$$ < $${\Phi _2}$$ = $${\Phi _3}$$ > $${\Phi _4}$$
$${\Phi _1}$$ > $${\Phi _2}$$ > $${\Phi _3}$$ > $${\Phi _4}$$
$${\Phi _1}$$ = $${\Phi _2}$$ = $${\Phi _3}$$ = $${\Phi _4}$$
$${\Phi _1}$$ > $${\Phi _3}$$ ; $${\Phi _2}$$ < $${\Phi _4}$$
Explanation
Net flux through a closed surface,
$$\phi $$ = $${{{q_{enclose}}} \over {{\varepsilon _0}}}$$
qenclosed = charge enclosed by closed surface.
For surface S1,
$$\phi $$1 = $${1 \over {{\varepsilon _0}}}$$ (2q)
For surface S2,
$$\phi $$2 = $${1 \over {{\varepsilon _0}}}$$ (q + q + q $$-$$ q) = $${1 \over {{\varepsilon _0}}}$$(2q)
For Surface S3,
$$\phi $$3 = $${1 \over {{\varepsilon _0}}}$$ (q + q) = $${1 \over {{\varepsilon _0}}}$$ (2q)
For surface S4,
$$\phi $$4 = $${1 \over {{\varepsilon _0}}}$$ (8q $$-$$ 2q $$-$$ 4q) = $${1 \over {{\varepsilon _0}}}$$ (2q)
$$\therefore\,\,\,$$ $$\phi $$1 = $$\phi $$2 = $$\phi $$3 = $$\phi $$4
$$\phi $$ = $${{{q_{enclose}}} \over {{\varepsilon _0}}}$$
qenclosed = charge enclosed by closed surface.
For surface S1,
$$\phi $$1 = $${1 \over {{\varepsilon _0}}}$$ (2q)
For surface S2,
$$\phi $$2 = $${1 \over {{\varepsilon _0}}}$$ (q + q + q $$-$$ q) = $${1 \over {{\varepsilon _0}}}$$(2q)
For Surface S3,
$$\phi $$3 = $${1 \over {{\varepsilon _0}}}$$ (q + q) = $${1 \over {{\varepsilon _0}}}$$ (2q)
For surface S4,
$$\phi $$4 = $${1 \over {{\varepsilon _0}}}$$ (8q $$-$$ 2q $$-$$ 4q) = $${1 \over {{\varepsilon _0}}}$$ (2q)
$$\therefore\,\,\,$$ $$\phi $$1 = $$\phi $$2 = $$\phi $$3 = $$\phi $$4
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