Question

Two thin conducting plates, each 24.0 cm on a side, are situated parallel to one another and 3.4 mm apart. If 1012 electrons are moved from one plate to the other, what is the electric field between the plates?

Answer 1

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

$E=\frac{(Q/A)}{E}\\ E=\frac{Q}{EA}\\ E=\frac{ne}{EA}\\ E=\frac{1012*()1.6*10^{-19} }{8.85*10^{-12}(0.24)(0.24)}\\ E=3.307*10^{-4}N/C$