A small charge that has lost 5.0 X 10^3 electrons and has a mass of 2.0 x 10^-6 kg experiences an acceleration of 15ms^2 when placed at a point P in space. Determine the magnitude and direction of the electric field at P

4 answers

15m/s^2
F = q E = m a

E = m a/q
m is given
q = 5*10^3 * electron charge
a is given
Why is 5*10^3 multiplied by a electron charge?
That is how much negative charge it is missing. (How much positive charge it has)