JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 8)
Charge is distributed within a sphere of radius R with a volume charge density $$\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},$$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :
a log $$\left( {1 - {Q \over {2\pi aA}}} \right)$$
$${a \over 2}$$ log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$
a log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$
$${a \over 2}$$ log $$\left( {1 - {Q \over {2\pi aA}}} \right)$$
Explanation
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Volume of this spherical layer,
dv = (4$$\pi $$r2)dr
charge present in this layer,
dq = $$\rho $$ (4$$\pi $$r2 dr)
= $${A \over {{r^2}}}{e^{ - {{2r} \over a}}}\,\,\left( {4\pi {r^2}dr} \right)$$
= $$A\,{e^{ - {{2r} \over a}}}\left( {4\pi dr} \right)$$
$$ \therefore $$ Total charge in the sphere,
Q= $$\int\limits_0^R {4\pi A{e^{ - {{2r} \over a}}}} \,dr$$
= 4$$\pi $$A$$\int\limits_0^R {{e^{ - {{2r} \over a}}}} \,dr$$
= 4$$\pi $$A$$\left[ {{{{e^{ - {{2r} \over a}}}} \over { - {2 \over a}}}} \right]_0^R$$
= 4$$\pi $$A $$\left( { - {a \over 2}} \right)\left( {{e^{ - {{2R} \over a}}} - 1} \right)$$
$$ \therefore $$ Q = 2$$\pi $$aA $$\left( {1 - {e^{ - {{2R} \over a}}}} \right)$$
$$ \Rightarrow $$ $${1 - {e^{ - {{2R} \over a}}}}$$ = $${Q \over {2\pi aA}}$$
$$ \Rightarrow $$ $${{e^{ - {{2R} \over a}}}}$$ = 1 $$-$$ $${Q \over {2\pi aA}}$$
$$ \Rightarrow $$ $${e^{{{2R} \over a}}}$$ = $${1 \over {1 - {Q \over {2\pi aA}}}}$$
$$ \Rightarrow $$ $${{2R} \over a} = \log \left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$
$$ \Rightarrow $$ R = $${a \over 2}$$ log $$\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$$
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