An insulating solid sphere of the radius R is charged in a non - uniform manner such that the volume charge density `rho=(A)/(r )`, where A is a positive constant and r is the distance from the centre. The potential difference between the centre and surface of the sphere is

Options

A. `(AR)/(8epsilon_(0))`
B. `(AR)/(4epsilon_(0))`
C. `(AR)/(epsilon_(0))`
D. `(AR)/(2epsilon_(0))`

Answer

To find the potential difference between the center and the surface of an insulating solid sphere with a non-uniform charge density given by \(\rho = \frac{A}{r}\), we can follow these steps:

### Step 1: Determine the total charge \(Q\) within the sphere
The volume charge density is given as:
\[ \rho(r) = \frac{A}{r} \]
To find the total charge \(Q\) within the sphere, we need to integrate the charge density over the volume of the sphere. The differential charge \(dQ\) in a spherical shell of radius \(r\) and thickness \(dr\) is given by:
\[ dQ = \rho(r) \cdot dV = \rho(r) \cdot (4\pi r^2 dr) \]
Substituting for \(\rho(r)\):
\[ dQ = \frac{A}{r} \cdot (4\pi r^2 dr) = 4\pi A r \, dr \]
Now, integrate \(dQ\) from \(0\) to \(R\) (the radius of the sphere):
\[ Q = \int_0^R 4\pi A r \, dr \]
Calculating the integral:
\[ Q = 4\pi A \left[\frac{r^2}{2}\right]_0^R = 4\pi A \cdot \frac{R^2}{2} = 2\pi A R^2 \]

### Step 2: Find the electric field \(E\) inside the sphere
Using Gauss's law, the electric field \(E\) at a distance \(r\) from the center (for \(r < R\)) is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \(Q_{\text{enc}}\) is the charge enclosed by a Gaussian surface of radius \(r\):
\[ Q_{\text{enc}} = 2\pi A r^2 \]
Thus, applying Gauss's law:
\[ E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} \implies E \cdot 4\pi r^2 = \frac{2\pi A r^2}{\epsilon_0} \]
Solving for \(E\):
\[ E = \frac{2\pi A r^2}{4\pi \epsilon_0 r^2} = \frac{A}{2\epsilon_0} \]

### Step 3: Calculate the potential difference \(V\) between the center and the surface
The potential difference \(V\) between the center (at \(r=0\)) and the surface (at \(r=R\)) can be calculated using:
\[ V = -\int_0^R E \, dr \]
Substituting for \(E\):
\[ V = -\int_0^R \frac{A}{2\epsilon_0} \, dr = -\frac{A}{2\epsilon_0} \left[r\right]_0^R = -\frac{A}{2\epsilon_0} (R - 0) = -\frac{AR}{2\epsilon_0} \]
The potential at the surface is higher than at the center, so the potential difference is:
\[ V = \frac{AR}{2\epsilon_0} \]

### Final Answer
The potential difference between the center and the surface of the sphere is:
\[ \frac{AR}{2\epsilon_0} \]