Question

What is the magnitude of the electric field at a point 0.0055 m from a 0.0025 C charge?

Answer 1

$E = 7.45x10^{11} N/C$

The intensity of the electric field (E) or simply electric field at a point is a vector quantity representing the electric force (F ) that acts over a charge, Q', located at that point.

$E=K\frac{Q}{r^{2}}$, where K is the constant of propotionality, Q a charge, and r the distance between a point and the charge.

$E = 9x10^{9} \frac{Nm^{2} }{C^{2} } \frac{0.0025C}{(0.0055m)^{2}}$

$E = 7.45x10^{11} N/C$