Respuesta :
Answer:
The system in the question is a capacitor. So,
[tex]C = \epsilon_0 \frac{A}{d} = (8.85\times 10^{-12}) \times \frac{2}{4\times 10^{-6}} = 4.425 \times 10^{-6} F[/tex]
[tex]V = \frac{Q}{C} = \frac{1 \times 10^{-6}}{4.425 \times 10^{-6}} = 0.225 V\\[/tex]
[tex]E = V/d = 225/(4\times 10^{-6}) = 56.25 \times 10^6 ~ V/m[/tex]
E is constant between the plates, so the E-field at a distance of 1mm from the positive plate is [tex]56.25 \times 10^6~ V/m.[/tex]
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, [tex]V = \int{E(r)} \, dr = \int {E} \, dr = Er[/tex]. 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.
