Question

Two 1.21 kg masses are 4.22 m apart on a frictionless table. Each has 22.78 microCoulombs of charge. What is the initial acceleration of each mass if they are released and allowed to move?

Answer 1

First, let's calculate the electric force, using the formula below:

$\begin{gathered} F_e=\frac{KQq}{d^2}\\ \\ F_e=\frac{9\cdot10^9\cdot(22.78\cdot10^{-6})^2}{4.22^2}\\ \\ F_e=262.26\cdot10^{-3}=0.26226\text{ N} \end{gathered}$

Now, let's use the second law of Newton to find the acceleration:

$\begin{gathered} F=m\cdot a\\ \\ 0.26226=1.21\cdot a\\ \\ a=\frac{0.26226}{1.21}\\ \\ a=0.217\text{ m/s^^b2} \end{gathered}$