Four point charges `-Q, -q, 2q` and `2Q` are placed, one at each corner of the square. The relation between `Q` and `q` for which the potential at the centre of the square is zero is

Options

A. `Q =- q`
B. `Q =- (1)/(q)`
C. `Q = q`
D. `Q = (1)/(q)`

Answer

To find the relation between the charges \( Q \) and \( q \) such that the potential at the center of the square formed by the charges is zero, we can follow these steps:

### Step 1: Identify the Configuration
We have four point charges placed at the corners of a square:
- Charge at corner 1: \( -Q \)
- Charge at corner 2: \( -q \)
- Charge at corner 3: \( 2q \)
- Charge at corner 4: \( 2Q \)

### Step 2: Determine the Distance from the Center
The distance from the center of the square to each corner can be calculated using the diagonal of the square. If the side length of the square is \( a \), the diagonal \( d \) is given by:
\[ d = a\sqrt{2} \]
The distance from the center to a corner (which is half the diagonal) is:
\[ r = \frac{d}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} \]

### Step 3: Calculate the Potential at the Center
The electric potential \( V \) at a point due to a point charge is given by:
\[ V = k \frac{q}{r} \]
where \( k \) is Coulomb's constant. The total potential at the center due to all four charges is the sum of the potentials due to each charge.

1. Potential due to \( -Q \):
\[ V_{-Q} = k \frac{-Q}{\frac{a}{\sqrt{2}}} = -k \frac{Q \sqrt{2}}{a} \]

2. Potential due to \( -q \):
\[ V_{-q} = k \frac{-q}{\frac{a}{\sqrt{2}}} = -k \frac{q \sqrt{2}}{a} \]

3. Potential due to \( 2q \):
\[ V_{2q} = k \frac{2q}{\frac{a}{\sqrt{2}}} = k \frac{2q \sqrt{2}}{a} \]

4. Potential due to \( 2Q \):
\[ V_{2Q} = k \frac{2Q}{\frac{a}{\sqrt{2}}} = k \frac{2Q \sqrt{2}}{a} \]

### Step 4: Sum the Potentials
Now, we sum the potentials at the center:
\[ V_{total} = V_{-Q} + V_{-q} + V_{2q} + V_{2Q} \]
Substituting the expressions we derived:
\[ V_{total} = -k \frac{Q \sqrt{2}}{a} - k \frac{q \sqrt{2}}{a} + k \frac{2q \sqrt{2}}{a} + k \frac{2Q \sqrt{2}}{a} \]

### Step 5: Factor Out Common Terms
Factoring out \( k \frac{\sqrt{2}}{a} \):
\[ V_{total} = k \frac{\sqrt{2}}{a} \left(-Q - q + 2q + 2Q\right) \]
This simplifies to:
\[ V_{total} = k \frac{\sqrt{2}}{a} \left(Q + q\right) \]

### Step 6: Set the Total Potential to Zero
For the potential at the center to be zero:
\[ k \frac{\sqrt{2}}{a} (Q + q) = 0 \]
Since \( k \frac{\sqrt{2}}{a} \) is not zero, we have:
\[ Q + q = 0 \]
Thus, we find:
\[ Q = -q \]

### Final Relation
The relation between \( Q \) and \( q \) for which the potential at the center of the square is zero is:
\[ Q = -q \]