JEE Advance - Physics (2020 - Paper 1 Offline - No. 4)
A circular disc of radius R carries surface charge density
$$\sigma \left( r \right) = {\sigma _0}\left( {1 - {r \over R}} \right)$$, where $$\sigma $$0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $$\phi $$0. Electric flux through another spherical surface of radius $${R \over 4}$$ and concentric with the disc is $$\phi $$. Then the ratio $${{{\phi _0}} \over \phi }$$ is ________.
$$\sigma \left( r \right) = {\sigma _0}\left( {1 - {r \over R}} \right)$$, where $$\sigma $$0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $$\phi $$0. Electric flux through another spherical surface of radius $${R \over 4}$$ and concentric with the disc is $$\phi $$. Then the ratio $${{{\phi _0}} \over \phi }$$ is ________.
Answer
6.4
Explanation
Let us consider a ring element of radius r and thickness dr.
Surface charge density of disc of radius R,
$$\sigma \left( r \right) = {\sigma _0}\left( {1 - {r \over R}} \right)$$
Charge of disc element, dq = $${\sigma _0}\left( {1 - {r \over R}} \right)2\pi rdr$$
$${\phi _0} = {{\int {dq} } \over {{\varepsilon _0}}} = {{\int\limits_0^R {{\sigma _0}\left( {1 - {r \over R}} \right)2\pi r\,dr} } \over {{\varepsilon _0}}}$$
$${\phi _0} = {{\int {dq} } \over {{\varepsilon _0}}} = {{\int\limits_0^{R/4} {{\sigma _0}\left( {1 - {r \over R}} \right)2\pi r\,dr} } \over {{\varepsilon _0}}}$$
$$ \therefore $$ $${{{\phi _0}} \over \phi } = {{{\sigma _0}2\pi \int\limits_0^R {\left( {r - {{{r^2}} \over R}} \right)} \,dr} \over {{\sigma _0}2\pi \int\limits_0^{R/4} {\left( {r - {{{r^2}} \over R}} \right)} \,dr}}$$
$$ = {{{{{R^2}} \over 2} - {{{R^2}} \over 3}} \over {{{{R^2}} \over {32}} - {{{R^2}} \over {3 \times 64}}}} = {{32} \over 5} = 6.40$$
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