Question 832: Electrostatics & Capacitance

Question

A charged shell of radius R carries a total charge Q. Given $$\phi $$ as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct?
[$$ \in $$
0
is the permittivity of free space]

Accepted Answer

(a) h > 2R and r > R
$$\phi  = {Q \over {{\varepsilon _0}}}$$, clearly from Gauss' Law
(b) suppose $$h = {{8R} \over 5}$$ and $$r = {{3R} \over 5}$$
$$ 	herefore $$ $$\phi $$ = 0
so, for $$h  2R and $$r = {{4R} \over 5}$$
$$ 	herefore $$ $$\phi  = {{2Q} \over {5{\varepsilon _0}}}$$
(d) like option c for h = 2R and $$r = {{3R} \over 5}$$
$${q_{enclosed}} = 2 	imes 2\pi (1 - \cos \,37^\circ ){Q \over {4\pi }} = {Q \over 5}$$
$$ 	herefore $$ Electric flux, $$\phi  = {Q \over {5{\varepsilon _0}}}$$

Answer

If h > 2R and r = 4R / 5 then $$\phi $$ = Q / 5 $$ \in $$
0

Answer

If h > 2R and r = 3R / 5 then $$\phi $$ = Q / 5 $$ \in $$
0

Answer

If h > 8R /5 and r = 3R / 5 then $$\phi $$ = 0

Answer

If h > 2R and r = R then $$\phi $$ = Q / $$ \in $$
0

Electrostatics & Capacitance

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