A system consists of a ball of radius R carrying a uniformly distributed charge q and the surrounding space filled with a charge of volume density `rho = alpha//r`, where `alpha` is a constant and r is the distance from the centre of the ball. Find the charge on the ball for which the magnitude of electric field strength outside the ball is constant? The dielectric constant of the ball and the surrounding may be taken equal to unity.

Explanation

To solve the problem, we need to find the charge \( q \) on a uniformly charged ball of radius \( R \) such that the electric field strength outside the ball remains constant. The surrounding space has a volume charge density given by \( \rho = \frac{\alpha}{r} \), where \( \alpha \) is a constant and \( r \) is the distance from the center of the ball.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a ball of radius \( R \) with charge \( q \).
- The surrounding space has a volume charge density \( \rho = \frac{\alpha}{r} \).
- We need to find \( q \) such that the electric field \( E \) outside the ball is constant.

2. **Applying Gauss's Law**:
- According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} \]

3. **Choosing a Gaussian Surface**:
- We choose a spherical Gaussian surface of radius \( r \) (where \( r > R \)) centered at the ball.
- The area of the Gaussian surface is \( 4\pi r^2 \).

4. **Calculating the Electric Field**:
- The electric field \( E \) is uniform over the surface, so the flux becomes:
\[ \Phi_E = E \cdot 4\pi r^2 \]

5. **Finding the Enclosed Charge \( Q_{\text{enc}} \)**:
- The total charge enclosed by the Gaussian surface is the sum of the charge on the ball and the charge from the surrounding volume.
- The charge \( Q_{\text{ball}} = q \).
- The charge from the surrounding volume can be calculated as:
\[ Q_{\text{surrounding}} = \int_{R}^{r} \rho \, dV = \int_{R}^{r} \left(\frac{\alpha}{r}\right) \, dV \]
- The differential volume element in spherical coordinates is \( dV = 4\pi r'^2 dr' \), where \( r' \) is the variable of integration.
- Therefore, we have:
\[ Q_{\text{surrounding}} = \int_{R}^{r} \frac{\alpha}{r'} (4\pi r'^2) dr' = 4\pi \alpha \int_{R}^{r} r' dr' = 4\pi \alpha \left[ \frac{r'^2}{2} \right]_{R}^{r} = 4\pi \alpha \left( \frac{r^2}{2} - \frac{R^2}{2} \right) \]

6. **Total Enclosed Charge**:
- Thus, the total enclosed charge is:
\[ Q_{\text{enc}} = q + 2\pi \alpha \left( r^2 - R^2 \right) \]

7. **Substituting into Gauss's Law**:
- Now substituting into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{q + 2\pi \alpha (r^2 - R^2)}{\epsilon_0} \]
- Rearranging gives:
\[ E = \frac{q + 2\pi \alpha (r^2 - R^2)}{4\pi \epsilon_0 r^2} \]

8. **Condition for Constant Electric Field**:
- For \( E \) to be constant, the term \( q + 2\pi \alpha (r^2 - R^2) \) must not depend on \( r \). This means the coefficient of \( r^2 \) must equal zero:
\[ 2\pi \alpha = 0 \implies q = -2\pi \alpha R^2 \]

9. **Final Result**:
- Therefore, the charge on the ball for which the electric field strength outside the ball is constant is:
\[ q = 2\pi \alpha R^2 \]