Electrostatics

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Question 168 of 311

Three infinitely long wires with linear charge density λ\lambda are placed along the x−axis,yx-a x i s, y-axis and z−z- axis respectively. Which of the following denotes an equipotential surface?

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Answer: (C) (x2+y2)(y2+z2)(z2+x2)=\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)= constant

Potential due to an infinite wire is V=2kλln⁡rV=2 k \lambda \ln r, where rr is the distance from the wire.

Taking the point in space P(x,y,z)P(x, y, z)
Distance from wire along xx-axis is rx=y2+z2r_x=\sqrt{y^2+z^2}
Distance from wire along yy-axis is ry=x2+z2r_y=\sqrt{x^2+z^2}
Distance from wire along zz-axis is rz=x2+y2r_z=\sqrt{x^2+y^2}
⇒\Rightarrow Potential at PP due to wire along xx-axis is
Vx=2kλln⁡rxV_x=2 k \lambda \ln r_x

Potential at PP due to wire along yy-axis is
Vy=2kλln⁡ryV_y=2 k \lambda \ln r_y

Potential at PP due to wire along zz-axis is
Vz=2kλln⁡rzV_z=2 k \lambda \ln r_z
⇒\Rightarrow Not potential at P=V=Vx+Vy+VzP=V=V_x+V_y+V_z
or V=2kλln⁡rx+2kλln⁡ry+2kλln⁡rz\quad V=2 k \lambda \ln r_x+2 k \lambda \ln r_y+2 k \lambda \ln r_z
i.e. V=2kλln⁡(rxryrz)V=2 k \lambda \ln \left(r_x r_y r_z\right)
or
$\begin{aligned}
& V=2 k \lambda \ln \left(\sqrt{y^2+z^2} \sqrt{z^2+x^2} \sqrt{x^2+y^2}\right) \\
& =k \lambda \ln \left(y^2+z^2\right)\left(z^2+x^2\right)\left(x^2+y^2\right)
\end{aligned}$
⇒\Rightarrow For equipotential surface
(x2+y2)(y2+z2)(z2+x2)= constant \left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=\text { constant }

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