Question

A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south.
What is the charge of object A?

Answer 1

Answer:

Charge, $q=2.78\times 10^{-10}\ C$

Explanation:

Given that,

Electric field, E = 40 N/C

At a point P located 0.250 m directly north of A, d = 0.25 m

We need to find the charge on the object A. The electric field at a distance of r is given by :

$E=\dfrac{kq}{r^2}$

$q=\dfrac{Er^2}{k}$

$q=\dfrac{40\times (0.25)^2}{9\times 10^9}$

$q=2.78\times 10^{-10}\ C$

So, the charge on object A is $q=2.78\times 10^{-10}\ C$. Hence, this is the required solution.