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

Question

The electrostatic potential in a charged spherical region of radius

r

varies as

V=a r^{3}+b

, where

a

and

b

are constants. The total charge in the sphere of unit radius is

\alpha \times \pi a \in_{\mathrm{o}}

. The value of

\alpha

is

\_\_\_\_

.
(permittivity of vacuum is

\epsilon_{0}

)

Quantrex Academy · Accepted Answer

Correct answer: B. -12

From V = ar^3 + b , the electric field is E = -dV/dr = -3ar^2 . By Gauss's law: E(r) · 4π r^2 = Q(r)/ _0 , so -3ar^2 × 4π r^2 = Q(r)/ _0 . This gives Q(r) = -12π ar^4 _0 . At r = 1 : Q(1) = -12π a _0 . Since Q(1) = α × π a _0 , we have α = -12

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