A parallel plate capacitor consists of two 4.0 cm x 4.0 cm electrodes spaced 2.0 mm apart. the electric field strength inside the capacitor is 1.0 x 10^6 n/c. what is the charge (in nC) of each electrode?
C = Q/V
and C = k epsilon A /d = 1*8.85×10^-12 *(0.4m)^2/0.002m = 7.1×10^-10F
V= Ed = 10^6N/C x 0.002m = 2000V
Q = CV = 7.1×10^-10F x 2000V = 1.42×10^-6C
should be (0.04m)^2, so the answer will be -8, not -6. A little late but anyone who reads this in the future should enjoy it haha