Question

What is the magnitude of the electric field of a proton at a distance of 50 micrometers? _____________ (in units of N/C)

Answer 1

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 $r=50\mu m=50\times 10^{-6}m$

We know that charge on proton $q=1.6\times 10^{-16}C$

Electric field due to point charge is given by $E=\frac{Kq}{r^2}$, here K is a constant which value is $9\times 10^9Nm^2/C^2$

So electric field $E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C$