Question

Two particles with electric charges Q and —3Q are separated by a distance of 1.2 m. (a) If Q = 4.5 C, what is the electric force between the two particles? (b) If Q —4.5 C, how does the answer change?

Answer 1

a) The electric force between two particles can be calculated as follows:

$$F_e=k\dfrac{Q_1Q_2}{d^2},$$ $$F_e=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{4.5\ C\cdot(-3\cdot4.5\ C)}{(1.2\ m)^2}=-3.8\cdot10^{11}\ N.$$

Since the two charges are opposite in sign they will attract each other, therefore, the direction of the electric force is attractive (also, the sign minus indicates that the direction is attractive).

b) The electric force between two particles can be calculated as follows:

$$F_e=k\dfrac{Q_1Q_2}{d^2},$$ $$F_e=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{-4.5\ C\cdot(-3)\cdot(-4.5\ C)}{(1.2\ m)^2}=-3.8\cdot10^{11}\ N.$$

Since the two charges are opposite in sign they will attract each other, therefore, the direction of the electric force is attractive (also, the sign minus indicates that the direction is attractive).

Answer:

a) $F_e=3.8\cdot10^{11}\ N,$ attractive.

b) $F_e=3.8\cdot10^{11}\ N,$ attractive.