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

Electric field at distance of 50 micrometer due to a proton is 0.576 N/C

Step-by-step explanation:

We have given distance where we have to find the electric field
`r=50\mu m=50* 10^(-6)m`

We know that charge on proton
`q=1.6* 10^(-16)C`

Electric field due to point charge is given by
`E=(Kq)/(r^2)`, here K is a constant which value is
`9* 10^9Nm^2/C^2`

So electric field
`E=(9* 10^9* 1.6* 10^(-16))/((50* 10^(-6))^2)=0.576N/C`