Question

A parallel-plate capacitor is formed from two 1.0{\rm cm} \times 1.0{\rm cm} electrodes spaced 2.8{\rm mm} apart. The electric field strength inside the capacitor is 1.0\times10^{6}{N/C}}
part a What is the charge (in {\rm nC}) on positive electrode?
part b What is the charge (in {\rm nC}) on negative electrode?

Answer 1

C=εε₀A/d
ε=1
ε₀=8.85 •10⁻¹² F/m
A=1•1=1 cm²=1•10⁻⁴ m²
C=q/U=q/Ed
ε₀A/d= q/Ed
q=ε₀AE=8.85 •10⁻¹²•1•10⁻⁴•1•10⁶=
=8.85•10⁻¹⁰ C=0.885 nC
a. +0.885 nC
b. -0.885 nC