Question

Three infinitely long wires with linear charge density \( \lambda \) are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

• A. \( (x + y)(y + z)(z + x) = \text{constant} \)
• B. \(xyz = \text{constant}\)
• C. \( (x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant} \)
• D. \( xy + yz + zx = \text{constant} \)

Hint:

In electrostatics, equipotential surfaces are where the electric potential is constant, and they often involve simple relationships between the distances from the source charges.

Answer 1

The Correct Option is C

This problem involves calculating the electric potential \(v\) generated by a system of charged wires. The core principle applied is that electric potential is the inverse of the line integral of the electric field.

1. Electric Potential Definition:
The electric potential \(v\) is formally defined as the negative line integral of the electric field \(\mathbf{E}\) along a path element \(d\mathbf{r}\):
\[ v = - \int \mathbf{E} \cdot d\mathbf{r} \]

2. Electric Field of a Single Charged Wire:
The electric field \(\mathbf{E}\) produced by an infinitely long charged wire with linear charge density \(\lambda\) at a radial distance \(r\) is given by:
\[ \mathbf{E} = \frac{2k\lambda}{r} \] Here, \(k\) represents Coulomb's constant.

3. Potential Due to a Single Wire:
By integrating the electric field formula, the potential \(v\) for a single wire is derived as:
\[ v = - \int \frac{2k\lambda}{r} dr = -2k\lambda \int \frac{1}{r} dr = -2k\lambda \ln{r} + C \] where \(C\) is the constant of integration.

4. Potential Due to Multiple Wires:
Consider three wires positioned such that the distances from the point of interest to each wire are \(r_1 = \sqrt{x^2 + y^2}\), \(r_2 = \sqrt{y^2 + z^2}\), and \(r_3 = \sqrt{z^2 + x^2}\). The total potential is the superposition of potentials from each individual wire:
\[ v = -2k\lambda \ln{\sqrt{x^2 + y^2}} - 2k\lambda \ln{\sqrt{y^2 + z^2}} - 2k\lambda \ln{\sqrt{z^2 + x^2}} + C \] (Note: The sign convention was adjusted to align with a potential reference point, as the original prompt's solution omitted negative signs.)

5. Simplification and Final Expression:
Combining the logarithmic terms yields:
\[ v = -2k\lambda \left( \ln{\sqrt{x^2 + y^2}} + \ln{\sqrt{y^2 + z^2}} + \ln{\sqrt{z^2 + x^2}} \right) + C \] \[ v = -2k\lambda \ln{\left( \sqrt{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} \right)} + C \]

6. Condition for Constant Potential \(v = c\):
If the potential \(v\) is constant, set to \(c\):
\[ -2k\lambda \ln{\left( \sqrt{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} \right)} + C = c \] Rearranging the equation:
\[ \ln{\left( \sqrt{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} \right)} = \frac{C - c}{2k\lambda} = C' \] Since \(C'\) is an arbitrary constant:
\[ \sqrt{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} = e^{C'} = C'' \] Squaring both sides:
\[ (x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = (C'')^2 \] where \(C''\) represents a new constant.

Final Conclusion:
The condition for a constant electric potential is expressed as:
\[ (x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant} \]