How strong is the electric field between the plates of a $0.80$-$\mu \textrm{F}$ air-gap capacitor if they are 2.0 mm apart and each has a charge of $62 \; \mu \textrm{C}$?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The charge on each plate of the capacitor is equal to the capacitance times the voltage and for a parallel plate capacitor. The voltage is going to be electric field times the separation between the plates so we can solve for electric field by dividing both sides by C d and we get electric field is the charge in each plates divided by capacitance times the separation between them. So that's 62 times ten to the minus six coulombs charge divided by 0.8 times ten to he minus six farads times two times ten to the minus three meter separating them and you get 3.9 times ten to the four Newton's per coulomb must be the electric field.