Six charges , `q_1 = +1` `mu`C , `q_2 = +3` `mu`C , `q_3 = +4` `mu`C , `q_4 = -2` `mu`C , `q_5 = -3` `mu`C and `q_6 = -3` `mu`C are placed on a sphee of radius 10 cm. The potential at centre of sphere is

Options

A. `27xx10^5` V
B. Zero
C. `2.7xx10^5` V
D. `0.27xx10^5 V`

Answer

To find the potential at the center of a sphere with given charges on its surface, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Charges**: We have six charges:
- \( q_1 = +1 \, \mu C \)
- \( q_2 = +3 \, \mu C \)
- \( q_3 = +4 \, \mu C \)
- \( q_4 = -2 \, \mu C \)
- \( q_5 = -3 \, \mu C \)
- \( q_6 = -3 \, \mu C \)

2. **Calculate the Total Charge**: The total charge \( Q_{net} \) on the sphere is the sum of all the individual charges:
\[ Q_{net} = q_1 + q_2 + q_3 + q_4 + q_5 + q_6 \]
Substituting the values:
\[ Q_{net} = (+1 \, \mu C) + (+3 \, \mu C) + (+4 \, \mu C) + (-2 \, \mu C) + (-3 \, \mu C) + (-3 \, \mu C) \]
\[ Q_{net} = 1 + 3 + 4 - 2 - 3 - 3 = 0 \, \mu C \]

3. **Use the Formula for Potential**: The potential \( V \) at the center of the sphere due to a charge \( Q \) at a distance \( r \) is given by:
\[ V = \frac{k \cdot Q_{net}}{r} \]
where \( k \) is Coulomb's constant (\( k \approx 9 \times 10^9 \, N \cdot m^2/C^2 \)) and \( r \) is the radius of the sphere. Here, \( r = 10 \, cm = 0.1 \, m \).

4. **Substitute the Values**: Since \( Q_{net} = 0 \):
\[ V = \frac{k \cdot 0}{0.1} = 0 \]

5. **Conclusion**: The potential at the center of the sphere is \( 0 \, V \).

### Final Answer
The potential at the center of the sphere is **0 V**.