A parallel-plate capacitor is formed from two 3.1 cm - diameter electrodes Part What is the charge (in nC) on each electrode? Express your answer using two significant figures spaced 1.7 mm apart. The electric fiold strength inside the capacitor is 3.0 x100 N/C

Answer: 2*10^-3 nC

Explanation: The capacitor of two plates is given by C = εoA/d where A and d, are the area and the separation of the plates, respectively.

We also know that C=Q/V and V=E*d (E=electric field between the plates)

From the above expression we have:

Q = εo*A*E = 8.85 ^10-12*7.54 * 10^-4*300 = 2*10^-3 nC