Four charges of `6 muC,2muC,-12muC and 4 muC` are placed at the corners of a square of side 1m. The square is in x-y plane and its centre is at origin. Electric potential due to these charges is zero everywhere or the line

Options

A. `x=y,z=0`
B. `z=0=z`
C. `x=0=y`
D. `x=z,y=0`

Answer

To determine the line along which the electric potential due to the given charges is zero, we can follow these steps:

### Step 1: Understand the Configuration
We have four charges located at the corners of a square with a side length of 1 meter. The charges are:
- \( Q_1 = 6 \, \mu C \) at (0.5, 0.5)
- \( Q_2 = 2 \, \mu C \) at (0.5, -0.5)
- \( Q_3 = -12 \, \mu C \) at (-0.5, -0.5)
- \( Q_4 = 4 \, \mu C \) at (-0.5, 0.5)

The center of the square is at the origin (0, 0).

### Step 2: Calculate the Total Charge
The total charge \( Q_{total} \) is calculated as follows:
\[ Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 = 6 \, \mu C + 2 \, \mu C - 12 \, \mu C + 4 \, \mu C = 0 \, \mu C \]
Since the total charge is zero, it indicates that there is a balance of positive and negative charges.

### Step 3: Electric Potential Formula
The electric potential \( V \) at a point due to a point charge is given by:
\[ V = k \cdot \frac{Q}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is being calculated.

### Step 4: Determine the Potential at a Point
To find the potential at a point \( P \) located at distance \( D \) from the center of the square (along the z-axis), we can express the potential due to each charge:
\[ V_P = k \left( \frac{Q_1}{D_1} + \frac{Q_2}{D_2} + \frac{Q_3}{D_3} + \frac{Q_4}{D_4} \right) \]
where \( D_1, D_2, D_3, D_4 \) are the distances from the point \( P \) to each of the charges.

### Step 5: Symmetry Consideration
Due to the symmetry of the square and the arrangement of charges, the distances from the point along the z-axis to each charge will be equal. Thus, we can denote the distance from the center to any corner of the square as \( D \):
\[ D_1 = D_2 = D_3 = D_4 = D \]

### Step 6: Setting Up the Equation
Substituting the charges and their distances into the potential equation, we get:
\[ V_P = k \left( \frac{6 \, \mu C}{D} + \frac{2 \, \mu C}{D} + \frac{4 \, \mu C}{D} - \frac{12 \, \mu C}{D} \right) \]
This simplifies to:
\[ V_P = k \cdot \frac{(6 + 2 + 4 - 12) \, \mu C}{D} = k \cdot \frac{0}{D} = 0 \]

### Step 7: Conclusion
The electric potential is zero at any point along the z-axis (where \( x = 0 \) and \( y = 0 \)), confirming that the electric potential due to these charges is zero everywhere along the line that is perpendicular to the square and passes through the origin.

### Final Answer
The electric potential due to these charges is zero everywhere along the z-axis.

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