An infinite plane of charge has a planar charge density of + 2.2 nC/m² and is located in the yz plane at x=0.A + 3.3 nC fixed point charge is located at x = + 1.0 m.At what position on the x-axis between x = 0.0 m and x = + 1.0 m is the electric potential a minimum?

Question

Quizplus · Accepted Answer

To find the position where electric potential is minimum, we need to find the point on the x-axis where the electric field is parallel to the x-axis. At this point, the work done in moving a test charge along the x-axis will be minimum.

The electric field from the infinite plane of charge is only in the x-direction, and has a magnitude of:

$E=\frac{\sigma}{2\epsilon_0}=\frac{2.2 	imes 10^{-9}}{2	imes 8.85	imes 10^{-12}}=495.5	ext{ N/C}$

The electric field from the point charge is in the direction towards the point charge and is given by:

$E=\frac{kq}{r^2}=\frac{9	imes 10^9 	imes 3.3	imes 10^{-9}}{1^2}=2.97	imes 10^9	ext{ N/C}$

For the electric field to be parallel to the x-axis, the y and z components of the electric field must cancel out. The y-component of the electric field due to the point charge is

$E_y=\frac{9	imes 10^9 	imes 3.3	imes 10^{-9}}{(x-1)^2}$

and the z-component of the electric field due to the plane of charge is

$E_z=\frac{2.2	imes 10^{-9}}{2	imes 8.85	imes 10^{-12}}\sin	heta=\frac{2.2	imes 10^{-9}}{2	imes 8.85	imes 10^{-12}}\frac{x}{\sqrt{x^2+1}}$

where $	heta$ is the angle between the electric field and the z-axis.

For the y and z components to cancel out, we must have

$E_y=E_z$

$\frac{9	imes 10^9 	imes 3.3	imes 10^{-9}}{(x-1)^2}=\frac{2.2	imes 10^{-9}}{2	imes 8.85	imes 10^{-12}}\frac{x}{\sqrt{x^2+1}}$

Solving for $x$ gives $x=0.51	ext{ m}$.

An Infinite Plane of Charge Has a Planar Charge Density