Question

A particle of charge 3.85\times 10^{-8}~\text{C}3.85×10​−8​​ C experiences a force of magnitude 7.65\times 10^{-6}~\text{N}7.65×10​−6​​ N when it is placed in a particular point in an electric field. What is the magnitude of the electric field at that point?

Answer 1

given

A particle of charge is

q = 3.85 x 10​⁻⁸​​ C  

a force of magnitude is

F = 7.65 x 10​^−6​​ N

the magnitude of the electric field at that point is E = ?

using equation for force is F = q E

E = F / q

= 7.65 x 10​^−6​​ N / 3.85 x 10​⁻⁸​​ C

E = 198.7012 N/C

so the magnitude of the electric field at that point is E = 198.7012 N/C