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{ = }(Kq_1q_2)/(r^2)`
`\begin{gathered} q_1\text{ = 3.7 }*10^(-6)C \\ q_2\text{ = 4.9 }*10^(-6)C \end{gathered}`

The distance r = 0.33 m

`k\text{ = }9\text{ }*10^9Nm^2/C^2`

Substituting these values into the formula above:

`F\text{ = }\frac{9\text{ }*10^9*3.7*10^(-6)*4.9*10^(-6)}{0.33^2}`
`\begin{gathered} F\text{ = }(163.17*10^(-3))/(0.1089) \\ F\text{ = }1498.35\text{ }*10^(-3) \\ F\text{ = }1.498N \end{gathered}`

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