A parallelplate capacitor is formed from two 40 cm diameter

A parallel-plate capacitor is formed from two 4.0 cm -diameter electrodes spaced 2.0 mm apart. The electric field strength inside the capacitor is 5.0 Times 10^6 N/C. What is the charge (in nC) on each electrode? Express your answer using two significant figures.

Solution

Q = CV

V = E*d

C = eo*A/d

using these three equations:

Q = eo*A*E = 8.85*10^-12*3.14*0.02^2*5.0*10^6

Q = 5.56*10^-8 C

Q = 55.6 nC

Let me know if you have any doubt.