Question

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charge density for each plate has magnitude 47.0 nC/m2, what is the magnitude of E⃗ in the region between the plates?

Answer 1

Answer:

$5.3\times 10^3 N/C$

Explanation:

We are given that

Distance between plates=d=2.2 cm=$2.2\times 10^{-2} m$

$1 cm=10^{-2} m$

$\sigma=47nC/m^2=47\times 10^{-9}C/m^2$

Using $1 nC=10^{-9} C$

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

$E=\frac{\sigma}{2\epsilon_0}$

$E_1=\frac{\sigma}{2\epsilon_0}$

$E_2=\frac{\sigma}{2\epsilon_0}$

$E=E_1+E_2$

$E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Substitute the values

$E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}$

$E=5.3\times 10^3 N/C$

Hence, the magnitude of E in the region between the plates=$5.3\times 10^3 N/C$