JEE PYQ: Electrostatics Question 53
Question 53 - 2019 (09 Jan Shift 2)
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:
(1) $a\log\left(1 - \frac{Q}{2\pi aA}\right)$
(2) $\frac{a}{2}\log\left(\frac{1}{1 - \frac{Q}{2\pi aA}}\right)$
(3) $a\log\left(\frac{1}{1 - \frac{Q}{2\pi aA}}\right)$
(4) $\frac{a}{2}\log\left(1 - \frac{Q}{2\pi aA}\right)$
Show Answer
Answer: (2)
Solution
$Q = \int_0^R \frac{A}{r^2}e^{-2r/a} \cdot 4\pi r^2 dr = 4\pi A \int_0^R e^{-2r/a}dr = 4\pi A \cdot \frac{-a}{2}[e^{-2r/a}]_0^R = 2\pi aA(1 - e^{-2R/a})$. Solving: $R = \frac{a}{2}\log\left(\frac{1}{1 - Q/(2\pi aA)}\right)$.
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