Answer to Question #157646 in Physics for Priya

Question

Two positively charged particles each having charge 20microC,are kept at a distance of 2.0m from each other.determine the electric field due to each charge. Show the electric field vectors on an appropriate diagram

Answer 1

The electric field due to each charge can be found from the Coulomb's Law:

$$F_{e}=k\dfrac{Q_1Q_2}{r^2},$$ $$F_e=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{(20\cdot10^{-6}\ C)^2}{(2.0\ m)^2}=0.9\ N.$$

Since the two charges are both positively charged they will repel each other, therefore, the direction of the electric force is repulsive (also, the sign plus indicates that the direction is repulsive). Also, $F_{e12}=F_{e21}=F_e$.