A hollow sphere of radius 0.1 m has a charge of `5 xx10^(-8) C`. The potential at a distance of 5 cm from the centre of the sphere is
`(1/(4pi epsi_(0))=9xx10^(9) Nm^(2) C^(-2))`

Options

A. 4000 V
B. 4500 V
C. 5000 V
D. 6000 V

Answer

To find the electric potential at a distance of 5 cm from the center of a hollow sphere with a charge, we can follow these steps:

### Step 1: Identify the given values
- Radius of the hollow sphere, \( R = 0.1 \, \text{m} \)
- Charge on the sphere, \( Q = 5 \times 10^{-8} \, \text{C} \)
- Distance from the center where we want to find the potential, \( r = 5 \, \text{cm} = 0.05 \, \text{m} \)

### Step 2: Understand the concept of electric potential
For a hollow sphere, the electric potential inside the sphere (at any point inside) is the same as the potential on the surface of the sphere. Thus, we can calculate the potential at the surface and use that value for the potential at 5 cm from the center.

### Step 3: Use the formula for electric potential
The formula for electric potential \( V \) at a distance \( r \) from a point charge is given by:
\[ V = \frac{kQ}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 4: Calculate the potential at the surface of the sphere
Since \( r = 0.05 \, \text{m} < R = 0.1 \, \text{m} \), we will calculate the potential at the surface of the sphere (at \( r = R = 0.1 \, \text{m} \)):
\[ V = \frac{kQ}{R} \]
Substituting the known values:
\[ V = \frac{9 \times 10^9 \times 5 \times 10^{-8}}{0.1} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ V = \frac{9 \times 10^9 \times 5 \times 10^{-8}}{0.1} = \frac{45 \times 10^1}{0.1} = 4500 \, \text{V} \]

### Conclusion
The potential at a distance of 5 cm from the center of the hollow sphere is \( 4500 \, \text{V} \).