What is the magnitude of the electric field of a proton at a distance of 25 micrometers?

Question

What is the magnitude of the electric field of a proton at a distance of 25 micrometers? 

Remember, the charge on a proton is 1.6e-19 C. (This is given on your formula sheet.)

 (in units of N/C)

Answer 1

Hello Arianna,

The electric field strength E due to a point charge q at a distance r is directly proportional to the charge and inversely proportional to the distance squared. The constant of proportionality k is 8.99 x 10^-9 N m² /C²

In equation form this is

E = k q/r²

Substituting the values given for this problem we get

E = (8.99 x 10⁹ N m²/C²) (1.6 x 10^-19 C)/(25 x 10^--6 m)² = 2.3 N/C

Hope this helps.

Best Regards,

Dean Creech

Answer 2

You can use the equation E = kQ / R^2 where k = 8.99 x 10^9 N m^2/C^2, R = 25 x 10^-6 m, C = 1.6 x 10^-19.C

E = 8.99 x 10^9 x 1.6 x 10^-19 / (25 x 10^-6)^2.

You can finish the calculation.