What is the magnitude of electrostatic force of attraction between two alpha particles separated

at a distance 5.3x10-11m? (1 apha particle = change of proton)

Answer: F = 3.28 * 10^-7 N

Explanation:

Alpha particle charge (q) = 2e

e = 1.6 * 10^-19

q = 2e = 2 * (1.6 * 10^-19) = 3.2 * 10^-19

1/4πEo = 9 * 10^9

Distance (r) = 5.3 * 10^-11m

The force of attraction between the two particles is given by:

F = (1/4πEo) (q1q2 / r^2)

F = [ (9 * 10^9) (3.2 * 10^-19) (3.2 * 10^-19) ] / (5.3 * 10^-11) ^2

F = (92.16 * 10^ (9 - 19 - 19)) / 28.09 * 10^-22

F = (92.16 * 10^-29) / 28.09 * 10^-22

F = 3.28 * 10^ (-29 + 22)

F = 3.28 * 10^-7 N