From V=ar3+b, the electric field is E=−dV/dr=−3ar2.
By Gauss's law:
E(r)⋅4πr2=Q(r)/ϵ0, so −3ar2×4πr2=Q(r)/ϵ0.
This gives Q(r)=−12πar4ϵ0.
At r=1: Q(1)=−12πaϵ0.
Since Q(1)=α×πaϵ0, we have α=−12
The electrostatic potential in a charged spherical region of radius r varies as V=ar3+b, where a and b are constants. The total charge in the sphere of unit radius is α×πa∈o. The value of α is ____.
(permittivity of vacuum is ϵ0)
Held on 24 Jan 2026 · Verified 6 Jul 2026.
−9
-12
-8
-6
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