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)

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

answer: option c.

explanation: