Question

Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electric force on one of the masses? What is the initial acceleration of each mass if they are released and allowed to move?

Answer 1

Force between two charges is given by

$F =\frac{kq_1q_2}{r^2}$

$F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}$

$F = 5.625 * 10^{-4} N$

Now in order to find the acceleration of each mass

we can use

F = ma

$5.625 * 10^{-4} = 1 * a$

$a= 5.625 * 10^{-4} m/s^2$

Answer 2

The force between two charged particles can be determined using the formula
F = k Q1 Q2 / d²

Using the given values
F = 9 x 10^9 (1.0 x 10^-9)(1.0 x 10^-9) / 4²
F = 5.625 x 10^-10 N

The acceleration is
ma = F
1 (a) = 5.625 x 10^-10
a = 5.625 x 10^-10 m/s²