Answer:
[tex]E=9.1\times 10^7 N/C[/tex]
Explanation:
We are given that
Length of side,l=6.4 cm=[tex]6.4\times 10^{-2} m[/tex]
1 m=100 cm
Total charge,q=[tex]6.6\times 10^{-6} C[/tex]
We have to find the magnitude E of the electric field just off the center of the plate.
Area charge density,[tex]\sigma=\frac{q}{A}=\frac{q}{l^2}=\frac{6.6\times 10^{-6}}{(6.4\times 10^{-2})^2}[/tex]
[tex]\sigma=16.1\times 10^{-4}C/m^2[/tex]
Electric field,[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Substitute the values
[tex]E=\frac{16.1\times 10^{-4}}{2\times 8.85\times 10^{-12}}[/tex]
[tex]E=9.1\times 10^7 N/C[/tex]
