JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2019
- Hard
- MCQ
Charge is distributed within a sphere of radius \(R\) with a volume charge density \(\rho (r) = \frac{A}{{{r^2}}}{e^{ - 2r/a}}\) where \(A\) and \(a\) are constants. If \(Q\) is the total charge of this charge distribution, the radius \(R\) is.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{a}{2}\,\log \,\left( {\frac{1}{{1 - \frac{Q}{{2\pi aA}}}}} \right)\)
Step-by-step Solution
Detailed explanation
\(Q = \int \rho 4\pi {r^2}dr = \int_0^R {\left( {\frac{A}{{{r^2}}}{e^{ - \frac{{2r}}{a}}}} \right)} \left( {4\pi {r^2}} \right)dr\)
\( = 4\pi A\frac{a}{2}\left( {1 - {e^{\frac{{ - 2R}}{a}}}} \right)\)
\( \Rightarrow R = \frac{{ - a}}{2}\log \left( {1 - \frac{Q}{{2\pi Aa}}} \right)\)
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