Answer:
E=3.307×10⁻⁴N/C
Explanation:
Given data
Length of the plate side L=24.0 cm =0.24 m
Distance between the plates d= 3.4 mm
Number of electron moves from one plate to others n=1012 electrons
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Electron charge e=-1.6×10⁻¹⁹C
To find
Electric field between the plates
Solution
E=σ/ε₀
[tex]E=\frac{(Q/A)}{E}\\ E=\frac{Q}{EA}\\ E=\frac{ne}{EA}\\ E=\frac{1012*()1.6*10^{-19} }{8.85*10^{-12}(0.24)(0.24)}\\ E=3.307*10^{-4}N/C[/tex]
