Problem — Force between +2 and -3 microcoulomb charges

Problem

Two point charges $q_1 = +2{,}0\ \mu\text{C}$ and $q_2 = -3{,}0\ \mu\text{C}$ are $0{,}25\ \text{m}$ apart. Calculate the magnitude, direction and sense of the force between the two charges.

Solution

Coulomb’s law with $k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2$, $r = 0{,}25\ \text{m}$:

$F = k\,\frac{|q_1|\,|q_2|}{r^2} = 8{,}99\cdot 10^9 \cdot \frac{(2{,}0\cdot 10^{-6})(3{,}0\cdot 10^{-6})}{(0{,}25)^2} = \frac{8{,}99\cdot 10^9 \cdot 6{,}0\cdot 10^{-12}}{0{,}0625} \approx 0{,}90\ \text{N}$

Direction and sense. The force is directed along the line joining the two charges. Since they have opposite sign, the interaction is attractive: each charge is pulled towards the other.

$\ev{F\approx 0{,}90\ \text{N, attractive, along the line joining them}}$

Links

Topics: Elettrostatica Concepts: Legge di Coulomb