Three infinitely long wires with linear charge density λ are placed along the x -axis, y -axis and z axis respectively. ​

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?

(1) \(x y+y z+z x=\) constant

(2) \((x+y)(y+z)(z+x)=\) constant

(3) \(\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)=\) constant

(4) \(\mathrm{xyz}=\) constant

Answer 1

Correct option is : (3) \(\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)=\) constant 

\(\mathrm{v}=-\int \vec{\mathrm{E}} \cdot \mathrm{d} \vec{\mathrm{r}}=\int \frac{2 \mathrm{k} \lambda}{\mathrm{r}} \mathrm{dr}=2 \mathrm{k} \lambda \ln \mathrm{r}+\mathrm{c}\)  

Net potential due to all wire

\( \mathrm{v}=2 \mathrm{k} \lambda \ln \sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}+2 \mathrm{k} \lambda \ln \sqrt{\mathrm{y}^{2}+\mathrm{z}^{2}}+2 \mathrm{k} \lambda \ln \sqrt{\mathrm{z}^{2}+\mathrm{x}^{2}}+\mathrm{c}\) 

for v = c

\(\sqrt{\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)}=c\) 

\(\therefore\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left(\mathrm{y}^{2}+\mathrm{z}^{2}\right)\left(\mathrm{z}^{2}+\mathrm{x}^{2}\right)=\mathrm{c}\) 

where c = constant

Answer 2

Correct Option is (3)

We have three infinitely long wires with charge density lambda () placed along the x-axis, y-axis, and z-axis. We need to determine the equation of an equipotential surface around these wires.

1. Potential Due to a Single Infinite Line Charge

The potential at a perpendicular distance r from an infinitely long charged wire is given by:

V = ( / 2 * pi * epsilon_0) * ln(r)

where r is the distance from the wire.

2. Total Potential at Any Point (x, y, z)

Each wire contributes to the total potential at any point (x, y, z):

• The wire along the x-axis contributes based on distance sqrt(y² + z²).
• The wire along the y-axis contributes based on distance sqrt(z² + x²).
• The wire along the z-axis contributes based on distance sqrt(x² + y²).

Adding these contributions:

V = ( / 2 * pi * epsilon_0) * [ ln(sqrt(y^2 + z^2)) + ln(sqrt(z^2 + x^2)) + ln(sqrt(x^2 + y^2)) ]

Using logarithmic properties:

**V = ( / 2 * pi * epsilon_0) * ln [ (x^2 + y^2) * (y^2 + z^2) * (z^2 + x^2) ]^(1/2) **

3. Equipotential Surface Condition

Since an equipotential surface means the potential is constant:

(x^2 + y^2) * (y^2 + z^2) * (z^2 + x^2) = constant

4. Matching with Given Options

Comparing this with the given answer choices, we see that Option (3) matches exactly:

(x^2 + y^2) * (y^2 + z^2) * (z^2 + x^2) = constant

Thus, the correct answer is:

Option (3)