P19S2017 – Four Infinite Sheets

The electric field created by an infinite sheet of charge with surface charge density $\sigma $ is perpendicular to the sheet and directed away from the sheet (if $\sigma >0$). Its magnitude is equal to

\boxed{E=\frac{\sigma }{2{\varepsilon }_0}=\frac{4\pi k\sigma }{2}=\frac{4\pi \cdot 9\times {10}^9\cdot {10}^{-5}}{4\pi \cdot 2}=4.5\times {10}^{-4}\ N/C}

The logical way to approach this question is to use the principle of superposition to determine the net electric field at point $\left(1\right)$ due to the four sheets of charge (feel free). However, before doing so, let us notice that the two plates on the left constitute a parallel-plate capacitor, as do the two plates on the right. Since we know that the electric field of a parallel- plate capacitor is confined to the gap between the plates, we conclude that the electric field at point $\left(1\right)$ is zero. Indeed, the electric field would be confined in the gap between the plates of the capacitor on the left and in the gap between the plates of the capacitor on the right, leaving the middle region with a zero net electric field.

Thus, the electric field $E_1$ at point $\left(1\right)$ is equal to

E_1=0\ \ \ \ N/C

and we conclude that the electric force acting on charge $q$ at point $\left(1\right)$ is therefore equal to

\boxed{F_1=0\ \ \ N}

Point $\left(2\right)$ is located in between the plates of the right parallel-plate capacitor and therefore the electric field points to the right and is equal to

E_2=\frac{\sigma }{{\varepsilon }_0}

We conclude that the electric force acting on charge $q$ at point $\left(2\right)$ is directed to the left ($q$ is negative) and has magnitude

\boxed{F_2=\left|q\right|E=2\times {10}^{-4}\cdot 9\times {10}^4=18\ N}