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(q_1q_2)/(d^2)`

By applying our values we get

`F_e=(9*10^9)((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=(0.118)/(0.967)=0.122(m)/(s^2)`

Our final acceleration is 0.122 m/s^2