QJEEM19Phe7bd: JEE Main 2019 Physics - Class 12 JEEM19Phe7bd Solution: Charge is distributed within a sphere of radius R with a volume …

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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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