Answer:
Electric field at distance of 50 micrometer due to a proton is 0.576 N/C
Explanation:
We have given distance where we have to find the electric field [tex]r=50\mu m=50\times 10^{-6}m[/tex]
We know that charge on proton [tex]q=1.6\times 10^{-16}C[/tex]
Electric field due to point charge is given by [tex]E=\frac{Kq}{r^2}[/tex], here K is a constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]
So electric field [tex]E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C[/tex]
