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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
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