Question

An electric field is set up between two parallel plates, each of area 2.0 m2, by putting 1.0 μC of charge on one plate and −1.0 μC of charge on the other. The plates are separated by 4.0 mm with their centers opposite each other, and the charges are distributed uniformly over the surface of the plates. What is the magnitude of the electric field between the plates at a distance of 1.0 mm from the positive plate, but not near the edges of the plates? (ϵ0=8.85×10−12 C2/N · m2)

Answer 1

Answer:

The system in the question is a capacitor. So,

$C = \epsilon_0 \frac{A}{d} = (8.85\times 10^{-12}) \times \frac{2}{4\times 10^{-6}} = 4.425 \times 10^{-6} F$

$V = \frac{Q}{C} = \frac{1 \times 10^{-6}}{4.425 \times 10^{-6}} = 0.225 V$

$E = V/d = 225/(4\times 10^{-6}) = 56.25 \times 10^6 ~ V/m$

E is constant between the plates, so the E-field at a distance of 1mm from the positive plate is $56.25 \times 10^6~ V/m.$

Explanation:

According to the information given in the question, we can find the capacitance of this system. However, we should find a way to calculate E-field from capacitance. The way to do that is to find the potential difference between the plates and then calculate E using their relationship with V.

Normally, we should differentiate V in order to find E. Since E is constant between the plates, $V = \int{E(r)} \, dr = \int {E} \, dr = Er$. This potential difference is calculated according to the distance between the plates, so we should divide V by 4mm to find E.

E is constant between parallel plate capacitors. This can be proven by the Gauss’ Law, however it is not necessary for this question.