Charge is distributed within a sphere of radius R with a volume charge density $\rho(r)=\frac{A}{r^{2}} e^{-2 r / a}$, where $A$ and a are constants. If $Q$ is the total charge of this charge distribution, the radius R is:

(B) $\frac{a}{2} \log \left(\frac{1}{1-\frac{Q}{2 \pi a A}}\right)$

(A) a $\log \left(1-\frac{\mathrm{Q}}{2 \pi \mathrm{aA}}\right)$

(C) $\frac{a}{2} \log \left(\frac{1}{1-\frac{Q}{2 \pi a A}}\right)$

(D) $\frac{a}{2} \log \left(1-\frac{1}{2 \pi a A}\right)$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Electrostatics

Charge is distributed within a sphere of radius R with a volume charge density

ρ (r) = \frac{A}{r ^{2}} e^{- 2 r / a}

, where

A

and a are constants. If

Q

is the total charge of this charge distribution, the radius R is:

Answer: (B)

Chop the sphere into thin shells of radius

r

and thickness

d r

; each holds charge

ρ (r) 4 π r^{2} d r

. Because

ρ = \frac{A}{r ^{2}} e^{- 2 r / a}

, the

r^{2}

cancels and

Q = \int_{0}^{R} \frac{A}{r ^{2}} e^{- 2 r / a} 4 π r^{2} d r = 4 π A \int_{0}^{R} e^{- 2 r / a} d r

. The integral gives

4 π A \cdot \frac{a}{2} (1 - e^{- 2 R / a}) = 2 π a A (1 - e^{- 2 R / a})

. Solving for

R

e^{- 2 R / a} = 1 - \frac{Q}{2 π a A}

, so

R = \frac{a}{2} ln (\frac{1}{1 - \frac{Q}{2 π a A}})

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option