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Electrostatics question

2019 · 9 Jan · Shift 2 · Q56
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Electrostatics question

2019 · 9 Jan · Shift 2 · Q56

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Charge is distributed within a sphere of radius R with a volume charge density ρ(r)=Ar2e−2rs,\rho \left( r \right) = {A \over {{r^2}}}{e^{ - {{2r} \over s}}},ρ(r)=r2A​e−s2r​, where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :
  1. A
    a log (1−Q2πaA)\left( {1 - {Q \over {2\pi aA}}} \right)(1−2πaAQ​)
  2. B
    a2{a \over 2}2a​ log (11−Q2πaA)\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)(1−2πaAQ​1​)
  3. C
    a log (11−Q2πaA)\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)(1−2πaAQ​1​)
  4. D
    a2{a \over 2}2a​ log (1−Q2πaA)\left( {1 - {Q \over {2\pi aA}}} \right)(1−2πaAQ​)
View written solutionFree

Correct answer: B

  1. Given volume charge density
ρ(r)=Ar2e−2ra\rho(r)=\frac{A}{r^2}e^{-\frac{2r}{a}}ρ(r)=r2A​e−a2r​

We need the total charge inside a sphere of radius RRR:

Q=∫Vρ(r) dVQ=\int_V \rho(r)\,dVQ=∫V​ρ(r)dV

For spherical symmetry,

dV=4πr2drdV=4\pi r^2drdV=4πr2dr

So,

Q=∫0RAr2e−2ra⋅4πr2 drQ=\int_0^R \frac{A}{r^2}e^{-\frac{2r}{a}}\cdot 4\pi r^2\,drQ=∫0R​r2A​e−a2r​⋅4πr2dr

The r2r^2r2 cancels:

Q=4πA∫0Re−2radrQ=4\pi A\int_0^R e^{-\frac{2r}{a}}drQ=4πA∫0R​e−a2r​dr
  1. Evaluate the integral
∫e−2radr=−a2e−2ra\int e^{-\frac{2r}{a}}dr=-\frac{a}{2}e^{-\frac{2r}{a}}∫e−a2r​dr=−2a​e−a2r​

Therefore,

Q=4πA[−a2e−2ra]0RQ=4\pi A\left[-\frac{a}{2}e^{-\frac{2r}{a}}\right]_0^RQ=4πA[−2a​e−a2r​]0R​ Q=4πA(−a2e−2Ra+a2)Q=4\pi A\left(-\frac{a}{2}e^{-\frac{2R}{a}}+\frac{a}{2}\right)Q=4πA(−2a​e−a2R​+2a​) Q=2πaA(1−e−2Ra)Q=2\pi aA\left(1-e^{-\frac{2R}{a}}\right)Q=2πaA(1−e−a2R​)
  1. Solve for RRR
Q2πaA=1−e−2Ra\frac{Q}{2\pi aA}=1-e^{-\frac{2R}{a}}2πaAQ​=1−e−a2R​ e−2Ra=1−Q2πaAe^{-\frac{2R}{a}}=1-\frac{Q}{2\pi aA}e−a2R​=1−2πaAQ​

Taking logarithm,

−2Ra=ln⁡(1−Q2πaA)-\frac{2R}{a}=\ln\left(1-\frac{Q}{2\pi aA}\right)−a2R​=ln(1−2πaAQ​) R=−a2ln⁡(1−Q2πaA)R=-\frac{a}{2}\ln\left(1-\frac{Q}{2\pi aA}\right)R=−2a​ln(1−2πaAQ​)

Using −ln⁡x=ln⁡(1/x)-\ln x=\ln(1/x)−lnx=ln(1/x),

R=a2ln⁡(11−Q2πaA)R=\frac{a}{2}\ln\left(\frac{1}{1-\frac{Q}{2\pi aA}}\right)R=2a​ln(1−2πaAQ​1​)
  1. Match with options

This is exactly Option B:

R=a2log⁡(11−Q2πaA)\boxed{R=\frac{a}{2}\log\left(\frac{1}{1-\frac{Q}{2\pi aA}}\right)}R=2a​log(1−2πaAQ​1​)​

So the correct answer is B.

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