Question

Topic: Electric Field

4.2: Plane z=10 m carries charge of \(\sf 20 nC/m^2\). What is the electric field intensity?

Answer 1

If i get yr question right, for an infinite plane of charge, with charge *surface* density \(\sigma\), at a fixed \(z\) in Cartesian, the electric field \(\mathbf E\) is:


\(\mathbf E = \dfrac{\sigma}{2 \varepsilon_o} \mathbf{e_z}\)

ie, it will be:

- this bit is hard (for me, anyways) to comprehend at first ..... constant, ie the same however far away from the plane you go

and

- this bit is easy .... perpendicular to the plane.

To be more precise, ie to get the sign right, as the field is the force on a unit positive charge, it really should be:

\(\mathbf E = \dfrac{\sigma}{2 \varepsilon_o} sgn ( z - 10) \mathbf{e_z}\)


In terms of derivation - the Gaussian Surface is your friend

Have a look here:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

Explains it way better than an ordinary apple

🍏