Question

A 5.0 c charge is 10 m from a small test charge. what is the magnitude of the electric field at the location of the test charge ?

Answer 1

Answer:

$4.50*10^8\frac{N}{C}$

Explanation:

The electric field is generated by a charge and represents the force exerted on a test charge, that is, the electric force per unit of charge. Therefore the equation for the electric field can be obtained from Coulomb's law.

$E=\frac{F}{q}\\E=\frac{kq}{r^2}\\E=\frac{5C*8.99*10^9\frac{Nm^2}{C^2}}{(10 m)^2}=4.50*10^8\frac{N}{C}$