A spaceship encounters a single plane of charged particles, with the charge per unit area equal...

Question:

A spaceship encounters a single plane of charged particles, with the charge per unit area equal to {eq}\sigma {/eq}. Find the magnitude and direction of the electric field a short distance above the plane.

Relativistic Electric Field

When traveling at speeds close to that of light our conventional notions regarding absolute lengths, time intervals, masses, etc will be proven false. None of these is absolute and depends on the frame of reference. Likewise, electric and magnetic field strengths transform from one frame to another. The electric charge is, however, an invariant quantity.

We know that in the rest frame of a single plane of uniform charge density {eq}\displaystyle {\sigma} {/eq} the electric field strength is {eq}\displaystyle {\frac{\sigma}{2 \epsilon_0}} {/eq} normal to the surface.

Suppose the space ship is moving parallel to the plane then the dimension of the plane parallel to the direction of motion is contracted by a factor of {eq}\displaystyle {\frac{1}{\gamma}} {/eq} where {eq}\displaystyle {\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}} {/eq}. Here {eq}\displaystyle {v} {/eq} is the speed of the spaceship.

But the charge remains invariant. Therefore the charge density is ramped up by the gamma factor. Hence the electric field component perpendicular to the sheet becomes ,

{eq}\displaystyle { E_{perpendicular}=\frac{\gamma \sigma}{2 \epsilon_0}} {/eq}.

Since there is no parallel component for either the electric field or magnetic field in the rest frame, there is no parallel component of the E-field in the spaceship frame either.

In case the spaceship is moving perpendicular to the sheet, there is no length contraction in the dimensions of the sheet. Therefore the charge density would remain the same and so would the field. In this case, the field strength would be {eq}\displaystyle {\frac{\sigma}{2 \epsilon_0}} {/eq} normal to the plane just as in the rest frame.