Question

The region between two concentric spheres of radii $'a'$ and $'b'$, respectively (see figure), has volume charge density $\rho = \frac{A}{r}$, where $A$ is a constant and $r$ is the distance from the centre. At the centre of the spheres is a point charge $Q$. The value of $A$ such that the electric field in the region between the spheres will be constant, is :

• A. $\frac{Q}{2 \pi a^2}$
• B. $\frac{Q}{2 \pi (b^2 - a^2)}$
• C. $\frac{Q}{ \pi (a^2 - b^2)}$
• D. $\frac{2 Q}{\pi a^2}$

Answer 1

The Correct Option is A

 To determine the constant electric field between two concentric spheres due to a volume charge density and a point charge at the center, we need to apply the principles of electrostatics. Here, we understand that the electric field in the region is influenced by both the point charge at the center and the volume charge density in the spherical shell.

Given:

• Radii of inner and outer spheres: \(a\) and \(b\), respectively.
• Volume charge density: \(\rho = \frac{A}{r}\), where \(A\) is a constant and \(r\) is the radial distance from the center.
• Point charge at the center: \(Q\)

To make the electric field constant in the region between the two spheres, we calculate the electric field using Gauss's Law. Gauss's Law states:

\(\oint \mathbf{E} \cdot d\mathbf{A} = \frac{1}{\epsilon_0} \cdot \left(Q_{\text{enclosed}}\right)\)

Step-by-Step Solution:

1. Consider a Gaussian surface of radius \(r\) where \(a \leq r \leq b\).
2. The charge enclosed by this Gaussian surface includes both the point charge \(Q\) and the charge due to the volume charge density:
3. Calculation of the enclosed charge due to volume density:
4. Total charge enclosed becomes:
5. Apply Gauss's Law to determine the electric field \(E\):
6. Simplifying gives:
7. For the electric field to be constant with respect to \(r\), the terms depending on \(r\) should equal zero. This is possible when:
8. The condition simplifies to \(A = \frac{Q}{2\pi a^2}\)

Hence, the correct value of \(A\) required to maintain a constant electric field in the region is \(\frac{Q}{2 \pi a^2}\).