Question

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of the electric force? The constant k = 9 x 109 N-m2/C2

Answer 1

Explanation:

Charges,$q_1=8\ \mu C=8\times 10^{-6}\ C$

$q_2=-5\ \mu C=-5\times 10^{-6}\ C$

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N$

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.