Question

Charge of uniform density (0.30 nC/m2) is distributed over the xy plane, and charge of uniform density (−0.40 nC/m2) is distributed over the yz plane. What is the magnitude of the resulting electric field at any point not in either of the two charged planes?

Answer 1

Answer: E = 39.54 N/C

Explanation: Electric field can be determined using surface charge density:

$E = \frac{\sigma}{2\epsilon_{0}}$

where:

σ is surface charge density

$\epsilon_{0}$ is permitivitty of free space ($\epsilon_{0} = 8.85.10^{-12}$$C^{2}/N.m^{2}$)

Calculating resulting electric field:

$E=E_{1} - E_{2}$

$E = \frac{\sigma_{1}-\sigma_{2}}{2\epsilon_{0}}$

$E = \frac{[0.3-(-0.4)].10^{-9}}{2.8.85.10^{-12}}$

$E=0.03954.10^{3}$

E = 39.54

The resulting Electric Field at any point is 39.54N/C.