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Electrostatics Part 3

Section: Physics

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Question : 14 of 80

Marks: +1, -0

The electrostatic potential in a charged spherical region of radius

r

V = a r^{3} + b

a

b

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

\alpha \times \pi a \in_{0}

\alpha

\_\_\_\_

. (permittivity of vacuum is

E_{0}

[24 jan 2026 Shift 1]

-12
-8
-9
-6

Solution:

Concept:

Use Gauss's law to relate electric flux to enclosed charge.

Explanation:

The electric field is the negative gradient of potential:

\vec{E} = -\frac{dV}{dr}

(radial).

Given

V = a r^3 + b

, differentiate:

E = -\frac{d}{dr}(a r^3 + b) = -3a r^2

.

At

r = 1

(unit radius),

E(1) = -3a

(magnitude).

Flux through sphere of radius 1:

\Phi_E = E \times \text{surface area} = (-3a) \times 4\pi (1)^2 = -12\pi a

.

By Gauss's law,

\Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0}

. So

Q_{\text{enclosed}} = \Phi_E \epsilon_0 = -12\pi a \epsilon_0

.

Given

Q_{\text{enclosed}} = \alpha \pi a \epsilon_0

. Comparing,

\alpha = -12

.

Answer:

\alpha = -12

(Option A).

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