Four pointed charges are placed on vertices of a square.If all the charges are brought to any one of the vertices.The change in potential at the centre of square due to this would be

Options

A. Dependent on polarity of charges
B. A non-zero constant
C. Zero
D. Infinite

Answer

To solve the problem, we need to determine the change in electric potential at the center of a square when four point charges located at the vertices are moved to one vertex.

### Step-by-Step Solution:

1. **Identify the Charges and Configuration**:
- Let the four charges be \( Q_1, Q_2, Q_3, \) and \( Q_4 \).
- The charges are initially located at the vertices of a square with side length \( a \).

2. **Calculate the Initial Potential at the Center**:
- The distance from each vertex of the square to the center is given by \( \frac{a}{\sqrt{2}} \).
- The potential \( V \) at the center due to a point charge \( Q \) is given by the formula:
\[ V = k \frac{Q}{r} \]
where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point where the potential is being calculated.
- Therefore, the total potential at the center due to all four charges is:
\[ V_{\text{initial}} = k \frac{Q_1}{\frac{a}{\sqrt{2}}} + k \frac{Q_2}{\frac{a}{\sqrt{2}}} + k \frac{Q_3}{\frac{a}{\sqrt{2}}} + k \frac{Q_4}{\frac{a}{\sqrt{2}}} \]
- Simplifying this, we have:
\[ V_{\text{initial}} = \frac{k \sqrt{2}}{a} (Q_1 + Q_2 + Q_3 + Q_4) \]

3. **Rearranging Charges to One Vertex**:
- Now, we move all charges \( Q_1, Q_2, Q_3, \) and \( Q_4 \) to one vertex of the square.
- The distance from this vertex to the center remains the same, \( \frac{a}{\sqrt{2}} \).

4. **Calculate the New Potential at the Center**:
- The potential at the center due to the charges now located at one vertex is:
\[ V_{\text{new}} = k \frac{Q_1}{\frac{a}{\sqrt{2}}} + k \frac{Q_2}{\frac{a}{\sqrt{2}}} + k \frac{Q_3}{\frac{a}{\sqrt{2}}} + k \frac{Q_4}{\frac{a}{\sqrt{2}}} \]
- This simplifies to:
\[ V_{\text{new}} = \frac{k \sqrt{2}}{a} (Q_1 + Q_2 + Q_3 + Q_4) \]

5. **Determine the Change in Potential**:
- Since both the initial and new potential expressions are identical:
\[ V_{\text{initial}} = V_{\text{new}} \]
- Therefore, the change in potential \( \Delta V \) is:
\[ \Delta V = V_{\text{new}} - V_{\text{initial}} = 0 \]

### Conclusion:
The change in potential at the center of the square when all charges are brought to one vertex is zero.