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Electrostatics question

2011 · Shift 0 · Q63
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Electrostatics question

2011 · Shift 0 · Q63

JEE MainPhysicsElectrostaticsMCQ+4 / −1
The electrostatic potential inside a charged spherical ball is given by ϕ=ar2+b\phi = a{r^2} + bϕ=ar2+b where rrr is the distance from the center and a,ba,ba,b are constants. Then the charge density inside the ball is:
  1. A
    −6aε0r- 6a{\varepsilon _0}r−6aε0​r
  2. B
    −24πaε0- 24\pi a{\varepsilon _0}−24πaε0​
  3. C
    −6aε0- 6a{\varepsilon _0}−6aε0​
  4. D
    −24πε0r- 24\pi {\varepsilon _0}r−24πε0​r
View written solutionFree

Correct answer: C

  1. Use Poisson’s equation

For electrostatics,

∇2ϕ=−ρε0\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}∇2ϕ=−ε0​ρ​

So,

ρ=−ε0∇2ϕ\rho = -\varepsilon_0 \nabla^2 \phiρ=−ε0​∇2ϕ
  1. Given potential
ϕ=ar2+b\phi = ar^2 + bϕ=ar2+b

Since the potential depends only on rrr, we use the Laplacian in spherical symmetry:

∇2ϕ=1r2ddr(r2dϕdr)\nabla^2 \phi = \frac{1}{r^2}\frac{d}{dr}\left(r^2 \frac{d\phi}{dr}\right)∇2ϕ=r21​drd​(r2drdϕ​)
  1. Differentiate step-by-step

First,

dϕdr=ddr(ar2+b)=2ar\frac{d\phi}{dr} = \frac{d}{dr}(ar^2+b)=2ardrdϕ​=drd​(ar2+b)=2ar

Then,

r2dϕdr=r2(2ar)=2ar3r^2\frac{d\phi}{dr}=r^2(2ar)=2ar^3r2drdϕ​=r2(2ar)=2ar3

Now differentiate again:

ddr(2ar3)=6ar2\frac{d}{dr}(2ar^3)=6ar^2drd​(2ar3)=6ar2

Therefore,

∇2ϕ=1r2(6ar2)=6a\nabla^2 \phi = \frac{1}{r^2}(6ar^2)=6a∇2ϕ=r21​(6ar2)=6a
  1. Find charge density

Using

ρ=−ε0∇2ϕ\rho = -\varepsilon_0 \nabla^2 \phiρ=−ε0​∇2ϕ

we get

ρ=−ε0(6a)=−6aε0\rho = -\varepsilon_0(6a) = -6a\varepsilon_0ρ=−ε0​(6a)=−6aε0​
  1. Match with options

The correct option is:

C: −6aε0\boxed{\text{C: } -6a\varepsilon_0}C: −6aε0​​
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