Question

Two 0.967 kg masses are 4.439 m apart on a frictionless table. Each has 16.074 microCoulombs of charge. What is the initial acceleration of each mass if they are released and allowed to move?

Answer 1

Firstly, we can write the equation for the electric force. It is:

$F_e=k\frac{q_1q_2}{d^2}$

By applying our values we get

$F_e=(9*10^9)\frac{(16.074*10^{-6})*(16.074*10^{-6})}{(4.439)^2}=0.118N$

Now, if we remind ourselves of Newton's law, we know that

$\vec{F}=m.\vec{a}$

We know the mass, and we know the Force, so we can find out the acceleration, this gives us:

$0.118=0.967*a$

Thus

$a=\frac{0.118}{0.967}=0.122\frac{m}{s^2}$

Our final acceleration is 0.122 m/s^2