Answer to Question #159831 in Physics for Haji

Question

What must be the distance between the point charge q 1 = 26.3C and q 2 = 38.2C in order to have a electrostatic attractive force between them has a magnitude of 4.89N.

Answer 1

From the Coulomb's Law, we have:

$$F=\dfrac{kQ_1Q_2}{r^2},$$ $$r=\sqrt{\dfrac{kQ_1Q_2}{F}},$$ $$r=\sqrt{\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot26.3\ C\cdot38.2\ C}{4.89\ N}}=1.36\cdot10^6\ m.$$