Question

10 pointsWhat is the electric force between a 3.7 x 10^-6 C charge and a 4.9 x10^-6 C charge that are separated by a distance of .33 m?(Don't forget your units and whether the force is attractive or repulsive!!)Your answer

Answer 1

Answer:

The electric force = 1.498N

It is an attractive force

Explanations:

The force between two charges q₁ and q₂ separated by a distance r is calculated using the coulomb's law equation

$F\text{ = }\frac{Kq_1q_2}{r^2}$$\begin{gathered} q_1\text{ = 3.7 }\times10^{-6}C \\ q_2\text{ = 4.9 }\times10^{-6}C \end{gathered}$

The distance r = 0.33 m

$k\text{ = }9\text{ }\times10^9Nm^2/C^2$

Substituting these values into the formula above:

$F\text{ = }\frac{9\text{ }\times10^9\times3.7\times10^{-6}\times4.9\times10^{-6}}{0.33^2}$$\begin{gathered} F\text{ = }\frac{163.17\times10^{-3}}{0.1089} \\ F\text{ = }1498.35\text{ }\times10^{-3} \\ F\text{ = }1.498N \end{gathered}$

Since the Force gotten is positive, it is an attractive force