Electrostatics

JEE Main level, single correct, medium

152 of 311

Question 152 of 311

The electrostatic potential in a charged spherical region of radius rr varies as V=ar3+bV=a r^{3}+b, where aa and bb are constants. The total charge in the sphere of unit radius is α×πa∈o\alpha \times \pi a \in_{\mathrm{o}}. The value of α\alpha is ____\_\_\_\_.
(permittivity of vacuum is ϵ0\epsilon_{0})

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Answer: (B) -12

From V=ar3+bV = ar^3 + b, the electric field is E=−dV/dr=−3ar2E = -dV/dr = -3ar^2.
By Gauss's law:
E(r)⋅4πr2=Q(r)/ϵ0E(r) \cdot 4\pi r^2 = Q(r)/\epsilon_0, so −3ar2×4πr2=Q(r)/ϵ0-3ar^2 \times 4\pi r^2 = Q(r)/\epsilon_0.
This gives Q(r)=−12πar4ϵ0Q(r) = -12\pi ar^4\epsilon_0.
At r=1r = 1: Q(1)=−12πaϵ0Q(1) = -12\pi a\epsilon_0.
Since Q(1)=α×πaϵ0Q(1) = \alpha \times \pi a\epsilon_0, we have α=−12\alpha = -12

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