Answer:
F = 91.27 N
Explanation:
Parameters given:
Charge of alpha particle, q = +2.0e = [tex]2.0 * 1.60217662 * 10^{-19}[/tex] = [tex]3.204 * 10^{-19} C[/tex]
Charge of gold nucleus, Q = +79e = [tex]79 * 1.60217662 * 10^{-19}[/tex] = [tex]1.266 * 10^{-17} C[/tex]
Distance between the alpha particle and gold nucleus, r = [tex]2.0 * 10^{-14} m[/tex]
Electric force is given as:
F = [tex]\frac{kqQ}{r^2}[/tex]
where k = Coulomb's constant
Therefore:
[tex]F = \frac{9 * 10^9 * 3.204 * 10^{-19} * 1.266 * 10^{-17}}{(2.0 * 10^{-14})^2} \\\\\\F = 91.27 N[/tex]
The electric force acting on the alpha particle when the alpha particle is [tex]2.0 * 10^{-14} m[/tex] from the gold nucleus is 91.27 N.
