Question

2d The figure above shows three very thin parallel planes of charge with equal separation 2 cm between them. The charge density of the middle plane is +12 nC/m2 (nc"nano Coulomb") and that of the two oute planes is -6nC/m2. Your answers for the electric field must include the correct sign. We choose positive electric field along the positive y-direction What is the electric field at point A? N/C Submit Answer Tries 0/2 What is the electric field at point B? N/C Submit Answer Tries 0/2 What is the electric field at point C? N/C Submit Answer Tries 0/2 What is the electric field at point D? N/C Submit Answer Tries 0/2

Answer 1

The upper plane produces an electric field $E=-\sigma /2\epsilon _{0}$ above it and a field $E=\sigma /2\epsilon _{0}$ below it taking into account the direction of the field also.

Similarly the middle plane produces an electric field $E=2\sigma /2\epsilon _{0}=\sigma /\epsilon _{0}$ above it and a field $E=-\sigma /\epsilon _{0}$ below it.

The lower plane produces a field $E=\sigma /2\epsilon _{0}$ above it and a field $E=-\sigma /2\epsilon _{0}$ below it.

Therefore by the principle of superposition ,

$E=-\sigma /2\epsilon _{0}+\sigma /\epsilon _{0}+\sigma /2\epsilon _{0}=\sigma /\epsilon _{0}=6*10^{-9}C/m^{2}/8.85*10^{-12}C^{2}/Nm^{2}=678 N/C.$ at point A.

$E=\sigma /2\epsilon _{0}+\sigma /\epsilon _{0}+\sigma /2\epsilon _{0}=2\sigma /\epsilon _{0}$ =1356 N/C at point B.

$E=\sigma /2\epsilon _{0}-\sigma /\epsilon _{0}+\sigma /2\epsilon _{0}=0$ N/C at point C.

$E=\sigma /2\epsilon _{0}-\sigma /\epsilon _{0}-\sigma /2\epsilon _{0}=-\sigma /\epsilon _{0}$ =-678 N/C at point D.