Question

Two 0.476 kg masses are 1.494 m apart on a frictionless table. Each has 19.39 microCoulombs of charge. What is the initial acceleration of each mass if they are released and allowed to move?

Answer 1

We will have the following:

We will have that the electrostatic force:

`EF=(kc_1c_2)/(d^2)`

Now, for all SI units we will have that the constant is almost exactly 9*19^9 m/farad; so the force on each mass will be:

`\begin{gathered} (9\ast10^9)(1\ast10^(-6))(1\ast10^(-6))/(1.494)^2=(1)/(166)N \\ \\ \approx6.02\ast10^(-3)N \end{gathered}`

Now, we find the acceleration; that is:

`\begin{gathered} (1)/(166)N=(0.476kg)\alpha\Rightarrow\alpha=(125)/(9877)m/s^2 \\ \\ \Rightarrow\alpha\approx0.0127m/s^2 \end{gathered}`

Now, we will have that if both charges have opposite signs then the acceleration of ach mass is approximately 0.0127 m/s^2 in the direction toward the other.

If both charges have the same sign, then the acceleration will be approximately 0.0127m/s^2 in the direction away from the other.