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

The problem asks for the electric field at an inside point ($r=r_1$) of an insulating solid sphere with a non-uniform volume charge density $\rho = \frac{A}{r}$, where $A$ is a positive constant and $r$ is the distance from the center.

We can use Gauss's Law to find the electric field.

1. Symmetry: The charge distribution is spherically symmetric, so the electric field will be radial and its magnitude will depend only on the distance from the center.

2. Gaussian Surface: To find the electric field at a point $r_1$ inside the sphere, we choose a spherical Gaussian surface of radius $r_1$ concentric with the solid sphere.

3. Enclosed Charge ($Q_{enc}$): The volume charge density is given by $\rho = \frac{A}{r}$. To find the total charge enclosed within the Gaussian sphere of radius $r_1$, 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 $r_1$, we integrate $dQ$ from $r=0$ to $r=r_1$: $Q_{enc} = \int_{0}^{r_1} 4\pi A r dr$ $Q_{enc} = 4\pi A \left[\frac{r^2}{2}\right]_{0}^{r_1}$ $Q_{enc} = 4\pi A \left(\frac{r_1^2}{2} - 0\right)$ $Q_{enc} = 2\pi A r_1^2$

4. Gauss's Law: According to Gauss's Law, the electric flux through a closed surface is equal to the total enclosed charge divided by $\epsilon_0$: $\oint \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0}$ For the spherical Gaussian surface, the electric field $\vec{E}$ is perpendicular to the surface and has a constant magnitude $E$ at every point on the surface. So, the integral simplifies to: $E (4\pi r_1^2) = \frac{Q_{enc}}{\epsilon_0}$ Substitute the value of $Q_{enc}$: $E (4\pi r_1^2) = \frac{2\pi A r_1^2}{\epsilon_0}$

5. Solve for E: $E = \frac{2\pi A r_1^2}{4\pi r_1^2 \epsilon_0}$ $E = \frac{A}{2\epsilon_0}$

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