Consider a thin spherical shell of radius R consisting of uniform surface charge density `sigma`. The electric field at a point of distance x from its centre and outside the shell is

Options

A. inversely proportional to `sigma`
B. directly proportional to `x^(2)`
C. directlr proportional to R
D. inversely proportional to `x^(2)`

Explanation

To find the electric field at a point outside a thin spherical shell with uniform surface charge density \(\sigma\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**: We have a thin spherical shell of radius \(R\) with a uniform surface charge density \(\sigma\). We need to find the electric field at a point located at a distance \(x\) from the center of the shell, where \(x > R\).

2. **Use Gauss's Law**: According to Gauss's law, the electric field \(E\) due to a symmetric charge distribution can be calculated using the formula:
\[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \(\Phi_E\) is the electric flux, \(Q_{\text{enc}}\) is the charge enclosed by the Gaussian surface, and \(\epsilon_0\) is the permittivity of free space.

3. **Determine the Enclosed Charge**: The total charge \(Q\) on the spherical shell can be expressed in terms of the surface charge density \(\sigma\) and the surface area of the sphere:
\[ Q = \sigma \cdot A = \sigma \cdot 4\pi R^2 \]

4. **Choose a Gaussian Surface**: For a point outside the shell (at distance \(x\)), we can choose a spherical Gaussian surface of radius \(x\). The electric field \(E\) will be uniform over this surface due to symmetry.

5. **Calculate the Electric Flux**: The electric flux through the Gaussian surface is given by:
\[ \Phi_E = E \cdot 4\pi x^2 \]

6. **Apply Gauss's Law**: Setting the electric flux equal to the charge enclosed divided by \(\epsilon_0\):
\[ E \cdot 4\pi x^2 = \frac{Q}{\epsilon_0} = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0} \]

7. **Solve for the Electric Field \(E\)**: Rearranging the equation gives:
\[ E = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0 \cdot 4\pi x^2} = \frac{\sigma R^2}{\epsilon_0 x^2} \]

8. **Conclusion**: The electric field at a distance \(x\) from the center of the spherical shell, where \(x > R\), is given by:
\[ E = \frac{\sigma R^2}{\epsilon_0 x^2} \]

### Final Result:
The electric field at a point outside the shell is inversely proportional to the square of the distance \(x\) from the center of the shell.