Q. The electrostatic potential in a charged spherical region of radius r varies as
\( V = ar^3 + b \), where \(a\) and \(b\) are constants. The total charge in the sphere
of unit radius is \( \alpha \times \pi a \varepsilon_0 \).
The value of \( \alpha \) is ____ .
(permittivity of vacuum is \( \varepsilon_0 \))
Explanation (Complete Step by Step Calculation)
Given electrostatic potential:
\[
V = ar^3 + b
\]
Electric field is related to potential by:
\[
E = -\frac{dV}{dr}
\]
Differentiate \(V\) with respect to \(r\):
\[
\frac{dV}{dr} = 3ar^2
\]
Hence electric field:
\[
E = -3ar^2
\]
Using Gauss’s law in differential form:
\[
\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}
\]
For spherical symmetry:
\[
\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{d}{dr}(r^2 E)
\]
Substitute \(E = -3ar^2\):
\[
r^2E = -3ar^4
\]
\[
\frac{d}{dr}(r^2E) = -12ar^3
\]
Therefore,
\[
\nabla \cdot \vec{E} = \frac{-12ar^3}{r^2} = -12ar
\]
Charge density:
\[
\rho = \varepsilon_0 (-12ar)
\]
Total charge inside unit radius:
\[
Q = \int_0^1 \rho \, dV
\]
Volume element:
\[
dV = 4\pi r^2 dr
\]
Substitute values:
\[
Q = \int_0^1 (-12a\varepsilon_0 r)(4\pi r^2)dr
\]
\[
Q = -48\pi a\varepsilon_0 \int_0^1 r^3 dr
\]
\[
\int_0^1 r^3 dr = \frac{1}{4}
\]
\[
Q = -48\pi a\varepsilon_0 \times \frac{1}{4}
\]
\[
Q = -12\pi a\varepsilon_0
\]
Comparing with \( Q = \alpha \pi a \varepsilon_0 \),
\[
\alpha = -12
\]