An alpha particle charge 2.0e is sent at high speed toward a gold nucleus charge 79e What is the electrical force acting on the alpha particle when it is 3.0 10-14 m from the gold nucleus?

Answer 1

Assuming a positive charge on the Alpha particle...

F = (Q1*Q2)/(r2)*K

F = (158e2)/(3*10-14)2*(8.988*109)

F = 40.4951176224... N = 40N

Edit....

F=ke(q1)(q2)

r^2

=(8.99x109 N*m2/C2) (158x1.60x10-19C)2 / (2.0X10-14m)2

=91N (the m2and C2 cancels out)

Answer 2

The electrical force acting on the alpha particle can be calculated using Coulomb's law. Given the charges of the alpha particle (2.0e) and the gold nucleus (79e), as well as the distance between them (3.0 x 10^-14 m), you can calculate the force using the formula:

[ F = \dfrac{k \cdot |q1 \cdot q2|}{r^2} ]

Where F is the electrical force, k is Coulomb's constant, q1 & q2 are the charges, and r is the distance. Plug in the values to find the force.