Question

An insulating solid sphere of radius $R$ is charged in a non-uniform manner such that volume charge density $\rho = \frac{A}{r}$, where $A$ is positive constant and $r$ is the distance from the center. Electric field at any inside point at $r=r_1$ is:

• A. $\frac{1}{4\pi\epsilon_0}\frac{4\pi A}{r_1}$
• B. $\frac{1}{4\pi\epsilon_0}\frac{A}{r_1}$
• C. $\frac{A}{\pi\epsilon_0}$
• D. $\frac{A}{2\epsilon_0}$

Answer 1

Answer: (D)

(4) $\frac{A}{2\epsilon_0}$

Answer 2

Concept: Gauss's Law is used to find the electric field due to symmetric charge distributions.

Gaussian Surface: Consider a concentric spherical Gaussian surface of radius $r_1$ ($r_1 < R$) inside the solid sphere.

Gauss's Law Statement: $\oint \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0}$

Symmetry: Due to spherical symmetry, the electric field $\vec{E}$ is radial and has a constant magnitude $E$ on the Gaussian surface. Thus, $\oint \vec{E} \cdot d\vec{S} = E \cdot (4\pi r_1^2)$.

Enclosed Charge ($Q_{enc}$): The volume charge density is non-uniform, $\rho = \frac{A}{r}$. To find the charge enclosed within the Gaussian surface, we integrate the charge density over the volume of the Gaussian sphere.
Consider a spherical shell of thickness $dr$ at a distance $r$ from the center. Its volume element is $dV = 4\pi r^2 dr$.
The charge in this shell is $dq = \rho dV = \left(\frac{A}{r}\right) (4\pi r^2 dr) = 4\pi A r dr$.
To find the total enclosed charge $Q_{enc}$ within the radius $r_1$, integrate $dq$ from $r=0$ to $r=r_1$:

$Q_{enc} = \int_{0}^{r_1} 4\pi A r dr$ $Q_{enc} = 4\pi A \left[\frac{r^2}{2}\right]_{0}^{r_1}$ $Q_{enc} = 4\pi A \left(\frac{r_1^2}{2} - 0\right)$ $Q_{enc} = 2\pi A r_1^2$

Applying Gauss's Law: Substitute $Q_{enc}$ into Gauss's Law:

$E (4\pi r_1^2) = \frac{2\pi A r_1^2}{\epsilon_0}$

Solving for E:

$E = \frac{2\pi A r_1^2}{4\pi r_1^2 \epsilon_0}$ $E = \frac{A}{2\epsilon_0}$

The electric field at any inside point at $r=r_1$ is $\frac{A}{2\epsilon_0}$.