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

The problem aims to determine the electric potential \(v\) due to a system of charged wires. The fundamental principle used is that the electric potential is the negative line integral of the electric field.

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

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

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

4. Potential Due to All Wires:
Assuming we have three wires located in such a way that the distances from the point of interest to each wire are given by \(r_1 = \sqrt{x^2 + y^2}\), \(r_2 = \sqrt{y^2 + z^2}\), and \(r_3 = \sqrt{z^2 + x^2}\), the total potential is the sum of the potentials from each 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: I've changed the sign here, as the solution in the prompt seems to have dropped the negative signs, or chosen a slightly different reference point for 0 potential.)

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

6. Setting \(v = c\): If \(v\) is constant (equal to \(c\)), then: \[ -2k\lambda \ln{\left( \sqrt{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} \right)} + C = c \] \[ \ln{\left( \sqrt{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} \right)} = \frac{C - c}{2k\lambda} = C' \] Since C' is some other constant: \[ \sqrt{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} = e^{C'} = C'' \] \[ (x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = (C'')^2 \] Where \(C''\) is another constant.

Final Conclusion:
Therefore, the condition for constant potential is:
\[ (x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant} \]

Answer 2

Step 1: Understand the given setup.
We have three infinitely long wires carrying uniform linear charge density \( \lambda \), placed along the x-axis, y-axis, and z-axis respectively. Each wire creates a cylindrical electric field around itself, and we need to find an equation that represents an equipotential surface due to all three wires together.

Step 2: Electric potential due to a single infinite charged wire.
The electric potential at a perpendicular distance \( r \) from an infinite line charge is given by: \[ V = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{1}{r}\right) \] or equivalently, \[ V \propto -\ln(r). \] Thus, for each wire, the potential depends only on the perpendicular distance from that wire.

Step 3: Calculate the potential due to all three wires.
- For the wire along the x-axis: distance from a point \( (x, y, z) \) is \( \sqrt{y^2 + z^2} \).
So potential \( V_1 \propto -\ln(\sqrt{y^2 + z^2}) \).
- For the wire along the y-axis: distance from a point \( (x, y, z) \) is \( \sqrt{z^2 + x^2} \).
So potential \( V_2 \propto -\ln(\sqrt{z^2 + x^2}) \).
- For the wire along the z-axis: distance from a point \( (x, y, z) \) is \( \sqrt{x^2 + y^2} \).
So potential \( V_3 \propto -\ln(\sqrt{x^2 + y^2}) \).

The total potential at point \( (x, y, z) \) is: \[ V = V_1 + V_2 + V_3 \propto -\ln\left[ (x^2 + y^2)(y^2 + z^2)(z^2 + x^2) \right]^{1/2}. \]

Step 4: Derive the equation for an equipotential surface.
For an equipotential surface, \( V = \text{constant} \). Since \( V \) depends logarithmically on the product of the squared distances: \[ (x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}. \]

Final Answer:
\[ \boxed{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}} \]