A parallel-plate capacitor is formed from two 7.2 cm-diameter electrodes spaced 1.8 mm apart. The electric field strength inside the capacitor is 3.0*106N/C. What is the magnitude of the charge (in nC) on each electrode?

Answer: 432 nC

Explanation: In order to calculate the charge, we know that the electric field corresponding for two parallel-plate is given by:

E=Q / (area*e0)

In our case E=3.0x10^6 N/C

Area = Pix (0,072 m) ^2

e0 vacuum permittivity 8,85x 10^-12