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Electrostatics question

2018 · 15 Apr · Shift 2 · Q45
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Electrostatics question

2018 · 15 Apr · Shift 2 · Q45

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A solid ball of radius R has a charge density ρ\rhoρ given by ρ\rhoρ=ρ\rhoρ o (1 −rR-\frac{r}{R}−Rr​) for 0 ≤\le≤ r ≤\le≤ R. The electric field outside the ball is :
  1. A
    ρoR3∈or2{{{\rho _o}{R^3}} \over {{ \in _o}{r^2}}}∈o​r2ρo​R3​
  2. B
    ρoR312∈or2{{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}12∈o​r2ρo​R3​
  3. C
    4ρoR33∈or2{{4{\rho _o}{R^3}} \over {3{ \in _o}{r^2}}}3∈o​r24ρo​R3​
  4. D
    3ρoR34∈or2{{3{\rho _o}{R^3}} \over {4{ \in _o}{r^2}}}4∈o​r23ρo​R3​
View written solutionFree

Correct answer: B

  1. Given charge density

For a solid sphere of radius RRR,

ρ(r)=ρ0(1−rR),0≤r≤R\rho(r)=\rho_0\left(1-\frac{r}{R}\right), \qquad 0\le r\le Rρ(r)=ρ0​(1−Rr​),0≤r≤R

We need the electric field outside the sphere, i.e. for r>Rr>Rr>R.

  1. Use Gauss's law

Outside a spherically symmetric charge distribution, the electric field is the same as that due to a point charge equal to the total charge enclosed:

E(r)=14πε0Qtotr2E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q_{\text{tot}}}{r^2}E(r)=4πε0​1​r2Qtot​​

So first we calculate QtotQ_{\text{tot}}Qtot​.

  1. Compute total charge in the sphere

A spherical shell of radius rrr and thickness drdrdr has volume

dV=4πr2 drdV=4\pi r^2\,drdV=4πr2dr

Hence,

dq=ρ(r) dV=ρ0(1−rR)4πr2drdq=\rho(r)\,dV=\rho_0\left(1-\frac{r}{R}\right)4\pi r^2 drdq=ρ(r)dV=ρ0​(1−Rr​)4πr2dr

Therefore,

Qtot=∫0Rdq=4πρ0∫0R(1−rR)r2 drQ_{\text{tot}}=\int_0^R dq =4\pi\rho_0\int_0^R \left(1-\frac{r}{R}\right)r^2\,drQtot​=∫0R​dq=4πρ0​∫0R​(1−Rr​)r2dr

Expand inside the integral:

Qtot=4πρ0∫0R(r2−r3R)drQ_{\text{tot}}=4\pi\rho_0\int_0^R \left(r^2-\frac{r^3}{R}\right)drQtot​=4πρ0​∫0R​(r2−Rr3​)dr

Now integrate:

∫0Rr2 dr=R33,∫0Rr3 dr=R44\int_0^R r^2\,dr=\frac{R^3}{3}, \qquad \int_0^R r^3\,dr=\frac{R^4}{4}∫0R​r2dr=3R3​,∫0R​r3dr=4R4​

So,

Qtot=4πρ0(R33−1R⋅R44)Q_{\text{tot}}=4\pi\rho_0\left(\frac{R^3}{3}-\frac{1}{R}\cdot\frac{R^4}{4}\right)Qtot​=4πρ0​(3R3​−R1​⋅4R4​) Qtot=4πρ0(R33−R34)=4πρ0R3(13−14)Q_{\text{tot}}=4\pi\rho_0\left(\frac{R^3}{3}-\frac{R^3}{4}\right) =4\pi\rho_0 R^3\left(\frac{1}{3}-\frac{1}{4}\right)Qtot​=4πρ0​(3R3​−4R3​)=4πρ0​R3(31​−41​) Qtot=4πρ0R3⋅112=πρ0R33Q_{\text{tot}}=4\pi\rho_0 R^3\cdot\frac{1}{12} =\frac{\pi\rho_0 R^3}{3}Qtot​=4πρ0​R3⋅121​=3πρ0​R3​
  1. Electric field outside

Now,

E(r)=14πε0Qtotr2=14πε0⋅πρ0R3/3r2E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q_{\text{tot}}}{r^2} =\frac{1}{4\pi\varepsilon_0}\cdot\frac{\pi\rho_0 R^3/3}{r^2}E(r)=4πε0​1​r2Qtot​​=4πε0​1​⋅r2πρ0​R3/3​

Simplify:

E(r)=ρ0R312ε0r2E(r)=\frac{\rho_0 R^3}{12\varepsilon_0 r^2}E(r)=12ε0​r2ρ0​R3​
  1. Match with options

This corresponds to:

ρ0R312ε0r2\boxed{\frac{\rho_0 R^3}{12\varepsilon_0 r^2}}12ε0​r2ρ0​R3​​

which is Option B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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