Question

a point charge of +5.00C is seperated from another point charge of -7.5C by a distance of 4cm apart. calculate the electrostatic force off attraction between the point charges

Answer 1

The electrostatic force of interaction between two point charges is calculated according to Coulomb's law: $F = k \frac{q_1 q_2}{r^2}$, where $k = 9 \cdot 10^9 \frac{N m^2}{C^2}$ is Coulomb's constant, $q_1, q_2$ are charges and $r$ - distance between them.

Hence, $F = \frac{9 \cdot 10^9 \frac{N m^2}{C^2} \cdot 5 C \cdot 7.5 C}{(0.04 m)^2} \approx 2.1 \cdot 10^{14} N$.