Question

Problem 20.29

A parallel-plate capacitor is formed from two 1.0 cm

Answer 1

Neglecting the details at the edges of the plates, a capacitor consisting of two parallel plates of surface area S separated by a small distance d has the following capacity (see first link).
C = epsilon_0 S/d
The charge on the negative electrode is -q
and the charge of the positive electrode is +q where:
q = C U
In this U is the potential difference U = d E. Therefore:
q = C E d = epsilon_0 S E , S = area
q = epsilon_0 (0.01 m)^2 (1.0 e(+6) V/m) = 8.854... e(-10) C
q = 0.8854 nC
q = 0.89 nC

+0.89 nC on +ve electrode and -0.89nC on -ve electrode