Two point charges of magnitude `+q and -q` are placed at `(-d//2,0,0) and (d//2,0,0)` are respectively. Find the equation of the euipotential surface where the potential is zero.

Options

A. `z^2 + (y-a)^2 = 2a`
B. `z^2 + (y-a)^2 = 27a^2//4`
C. `z^2+y^2 = 15a^2//4`
D. none of these

Explanation

To find the equation of the equipotential surface where the potential is zero due to two point charges \( +q \) and \( -q \) placed at \( (-\frac{d}{2}, 0, 0) \) and \( (\frac{d}{2}, 0, 0) \) respectively, we can follow these steps:

### Step 1: Identify the Coordinates of the Charges
The positive charge \( +q \) is located at \( A(-\frac{d}{2}, 0, 0) \) and the negative charge \( -q \) is located at \( B(\frac{d}{2}, 0, 0) \).

### Step 2: Define the Point of Interest
Let \( P(x, y, z) \) be an arbitrary point in space where we want to find the potential.

### Step 3: Calculate Distances from Point P to Each Charge
Using the distance formula, we can find the distances from point \( P \) to each charge:
- Distance \( AP \) from \( A \) to \( P \):
\[ AP = \sqrt{(x + \frac{d}{2})^2 + y^2 + z^2} \]
- Distance \( BP \) from \( B \) to \( P \):
\[ BP = \sqrt{(x - \frac{d}{2})^2 + y^2 + z^2} \]

### Step 4: Write the Expression for Electric Potential
The electric potential \( V \) at point \( P \) due to the two charges is given by:
\[ V_P = V_A + V_B = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{AP} - \frac{q}{BP} \right) \]
Setting this equal to zero for the equipotential surface:
\[ \frac{q}{AP} - \frac{q}{BP} = 0 \]

### Step 5: Simplify the Equation
Since \( q \) is non-zero, we can simplify the equation:
\[ \frac{1}{AP} = \frac{1}{BP} \]
This leads to:
\[ AP = BP \]

### Step 6: Substitute the Distances
Substituting the expressions for \( AP \) and \( BP \):
\[ \sqrt{(x + \frac{d}{2})^2 + y^2 + z^2} = \sqrt{(x - \frac{d}{2})^2 + y^2 + z^2} \]

### Step 7: Square Both Sides
Squaring both sides to eliminate the square roots:
\[ (x + \frac{d}{2})^2 + y^2 + z^2 = (x - \frac{d}{2})^2 + y^2 + z^2 \]

### Step 8: Cancel Common Terms
The \( y^2 \) and \( z^2 \) terms cancel out:
\[ (x + \frac{d}{2})^2 = (x - \frac{d}{2})^2 \]

### Step 9: Expand Both Sides
Expanding both sides:
\[ x^2 + dx + \frac{d^2}{4} = x^2 - dx + \frac{d^2}{4} \]

### Step 10: Simplify the Equation
Cancelling \( x^2 \) and \( \frac{d^2}{4} \) from both sides:
\[ dx = -dx \]
This implies:
\[ 2dx = 0 \quad \Rightarrow \quad x = 0 \]

### Step 11: Conclusion
Since \( x = 0 \), the equipotential surface where the potential is zero is along the y-axis and z-axis:
\[ \text{The equation of the equipotential surface is } x = 0. \]