The maximum surface charge density of a uniformly charged spherical radius R and wall thickness 't', so that it will not burst apart, is

Options

A. `sqrt((sigma_m epsilon_0 t)/R)`
B. `sqrt((2sigma_m epsilon_0 t)/R)`
C. `sqrt((3sigma_m epsilon_0 t)/R)`
D. `sqrt((4sigma_m epsilon_0 t)/R)`

Answer

To find the maximum surface charge density (\( \sigma \)) of a uniformly charged spherical shell with radius \( R \) and wall thickness \( t \) so that it does not burst apart, we can follow these steps:

### Step 1: Understand the Forces Acting on the Shell
When the spherical shell is uniformly charged, each hemisphere experiences an attractive force due to the other hemisphere. This force can lead to stress on the shell.

### Step 2: Calculate the Charge on Each Hemisphere
Let \( Q \) be the total charge on one hemisphere. The surface charge density \( \sigma \) is defined as the charge per unit area. The area of one hemisphere is given by:
\[ A = 2\pi R^2 \]
Thus, the charge \( Q \) on one hemisphere can be expressed as:
\[ Q = \sigma \cdot A = \sigma \cdot 2\pi R^2 \]

### Step 3: Calculate the Force Between Hemispheres
The attractive force \( F \) between the two hemispheres can be calculated using Coulomb's law:
\[ F = k \frac{Q^2}{r^2} \]
where \( k \) is the Coulomb's constant and \( r \) is the distance between the centers of mass of the two hemispheres, which is equal to \( R \).

Substituting \( Q \) into the force equation:
\[ F = k \frac{(\sigma \cdot 2\pi R^2)^2}{R^2} = k \frac{4\pi^2 \sigma^2 R^4}{R^2} = 4\pi^2 k \sigma^2 R^2 \]

### Step 4: Relate the Force to Maximum Stress
The maximum stress \( \sigma_m \) that the material can withstand before bursting is given by:
\[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \]
The area on which this force acts is the area of the shell, which can be approximated as:
\[ A = 2\pi R \cdot t \]
Thus, the stress can be expressed as:
\[ \sigma_m = \frac{F}{A} = \frac{4\pi^2 k \sigma^2 R^2}{2\pi R t} = \frac{2\pi k \sigma^2 R}{t} \]

### Step 5: Solve for Maximum Surface Charge Density
Rearranging the equation to solve for \( \sigma \):
\[ \sigma^2 = \frac{\sigma_m t}{2\pi k R} \]
Taking the square root gives:
\[ \sigma = \sqrt{\frac{\sigma_m t}{2\pi k R}} \]

### Step 6: Substitute \( k \) with \( \frac{1}{4\pi \epsilon_0} \)
Substituting \( k \) into the equation:
\[ \sigma = \sqrt{\frac{\sigma_m t}{2\pi \cdot \frac{1}{4\pi \epsilon_0} R}} = \sqrt{\frac{4\pi^2 \sigma_m \epsilon_0 t}{2R}} = \sqrt{\frac{2\sigma_m \epsilon_0 t}{R}} \]

### Final Result
Thus, the maximum surface charge density is given by:
\[ \sigma = \sqrt{\frac{2\sigma_m \epsilon_0 t}{R}} \]