Surface charge density of a sphere of a radius 10 cm is `8.85 xx 10^(-8)c//m^(2)`. Potential at the centre of the sphere is

Options

A. 1000 V
B. 885 V
C. `10^(-3)` V
D. 442.5 V

Answer

To find the potential at the center of a sphere with a given surface charge density, we will follow these steps:

### Step 1: Identify the given values
- Radius of the sphere, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Surface charge density, \( \sigma = 8.85 \times 10^{-8} \, \text{C/m}^2 \)

### Step 2: Calculate the total charge \( Q \) on the sphere
The total charge \( Q \) on the sphere can be calculated using the formula:
\[ Q = \sigma \times A \]
where \( A \) is the surface area of the sphere given by \( A = 4\pi r^2 \).

Calculating the surface area:
\[ A = 4\pi (0.1)^2 = 4\pi (0.01) = 0.04\pi \, \text{m}^2 \]

Now substituting the values:
\[ Q = \sigma \times 0.04\pi = (8.85 \times 10^{-8}) \times (0.04\pi) \]

### Step 3: Calculate the potential \( V \) at the center of the sphere
The potential \( V \) at the center of the sphere is given by:
\[ V = k \frac{Q}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) and \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

Substituting \( Q \) into the potential formula:
\[ V = k \frac{\sigma \times 4\pi r^2}{r} \]
This simplifies to:
\[ V = k \sigma \times 4\pi r \]

### Step 4: Substitute the value of \( k \)
Substituting \( k \):
\[ V = \left(\frac{1}{4\pi \epsilon_0}\right) \sigma \times 4\pi r \]
The \( 4\pi \) cancels out:
\[ V = \frac{\sigma r}{\epsilon_0} \]

### Step 5: Substitute the known values
Now substituting the values of \( \sigma \), \( r \), and \( \epsilon_0 \):
\[ V = \frac{(8.85 \times 10^{-8}) \times (0.1)}{8.85 \times 10^{-12}} \]

### Step 6: Simplify the expression
\[ V = \frac{8.85 \times 10^{-9}}{8.85 \times 10^{-12}} = 10^3 \, \text{V} \]

### Final Answer
Thus, the potential at the center of the sphere is:
\[ V = 1000 \, \text{V} \]