Question

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 $a \times \pi \varepsilon_{0}$. The value of $a$ is ___.
(Permittivity of vacuum is $\varepsilon_{0}$)

• A. $-8$
• B. $-12$
• C. $-9$
• D. $-6$

Hint:

Electric field can always be obtained by taking the negative gradient of electrostatic potential.

Answer 1

The Correct Option is B

To find the value of \(a\), we start with the given electrostatic potential function inside a charged spherical region of radius \(r\):

\(V = ar^3 + b\)

We need to determine the electric field \(E\), which is related to the potential \(V\) by the relation:

\(E = -\nabla V\)

In spherical coordinates, only the radial component is considered for symmetry, therefore:

\(E = -\frac{dV}{dr} = -\frac{d}{dr}(ar^3 + b)\)

Calculating the derivative, we get:

\(E = - (3ar^2)\)

According to Gauss's Law, the total electric flux through a spherical surface of radius \(r\) is given by:

\(\Phi = E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0}\)

Where \(Q_{\text{enc}}\) is the charge enclosed within the radius \(r\). Substituting \(E\) and rearranging gives:

\(-3ar^2 \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0}\)

This simplifies to:

\(-12\pi a r^4 = \frac{Q_{\text{enc}}}{\varepsilon_0}\)

The total charge inside a unit sphere across the whole volume can be found by integrating:

\(Q_{\text{total}} = \int_0^{1}\rho \cdot 4\pi r^2 dr\)

Since \(\rho\) (the charge density) is related to the divergence of the electric field as per Gauss's law, consider:

\(\rho = -\frac{1}{r^2} \frac{d}{dr}(r^2 E) = -\frac{1}{r^2} \frac{d}{dr}(r^2(-3ar^2))\)

This evaluates to:

\(\rho = 12a\varepsilon_0r\)

The charge within the sphere of unit radius is:

\(Q_{\text{total}} = \int_0^1 12a\varepsilon_0 r \cdot 4\pi r^2 dr\)

Simplifying, we have:

\(Q_{\text{total}} = 48\pi a\varepsilon_0 \int_0^1 r^3 dr = 48\pi a\varepsilon_0 \cdot \frac{1}{4} = 12\pi a\varepsilon_0\)

Given that \(Q_{\text{total}} = a \cdot \pi\varepsilon_0\), comparing both expressions:

\(12\pi a\varepsilon_0 = a \cdot \pi\varepsilon_0\)

Simplifying this gives:

\(12a = a\)

Therefore, \(a\) must be:

-12

Thus, the correct value of \(a\) is -12.