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Question

(C) The potential $V$ at a distance $r$ from an infinitely long charged wire is given by $V = -\int \vec{E} \cdot d\vec{r} = -\int \frac{2k\lambda}{r}

Vedclass · Accepted Answer

$(x^2+y^2)(y^2+z^2)(z^2+x^2) = 	ext{constant}$

Vedclass · Answer

(C) The potential $V$ at a distance $r$ from an infinitely long charged wire is given by $V = -\int \vec{E} \cdot d\vec{r} = -\int \frac{2k\lambda}{r} dr = -2k\lambda \ln r + C$.For the three wires placed along the $x, y,$ and $z$ axes,the distances from the wires to a point $(x, y, z)$ are $r_x = \sqrt{y^2+z^2}$,$r_y = \sqrt{x^2+z^2}$,and $r_z = \sqrt{x^2+y^2}$ respectively.The net potential $V$ is the sum of the potentials due to each wire:$V = -2k\lambda \ln(\sqrt{y^2+z^2}) - 2k\lambda \ln(\sqrt{x^2+z^2}) - 2k\lambda \ln(\sqrt{x^2+y^2}) + C$.$V = -k\lambda \ln[(y^2+z^2)(x^2+z^2)(x^2+y^2)] + C$.For an equipotential surface,$V = 	ext{constant}$,which implies that the product $(x^2+y^2)(y^2+z^2)(z^2+x^2)$ must be constant.

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?

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