An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electric force acting on the alpha particle when the alpha particle is 1.7 multiplied by 10-14 m from the gold nucleus?
N
coulombs law...
F=kq1*q2/r^2
To calculate the electric force acting on the alpha particle, we can use Coulomb's Law. Coulomb's Law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * (q1 * q2) / r^2
Where:
F = Electric force
k = Coulomb's constant = 8.99 x 10^9 Nm^2/C^2
q1, q2 = Charges of the two objects
r = Distance between the two objects
In this case, the charge of the alpha particle is +2.0e, which is equivalent to +2.0 * 1.6 x 10^-19 C. The charge of the gold nucleus is +79e, which is equivalent to +79 * 1.6 x 10^-19 C. The distance between them is 1.7 x 10^-14 m.
Substituting the values into the equation:
F = (8.99 x 10^9 Nm^2/C^2) * [(2.0 * 1.6 x 10^-19 C) * (79 * 1.6 x 10^-19 C)] / (1.7 x 10^-14 m)^2
Calculating the value:
F ≈ 5.49 x 10^-11 N
Therefore, the electric force acting on the alpha particle when it is 1.7 x 10^-14 m from the gold nucleus is approximately 5.49 x 10^-11 N.
To find the electric force acting on the alpha particle, we can use Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's Law Formula:
F = k * (q1 * q2) / r^2
Where:
F is the electric force
k is the electrostatic constant (9 * 10^9 N m^2 / C^2)
q1 and q2 are the charges of the two particles
r is the distance between the particles
In this case, the charge of the alpha particle (q1) is +2.0e (where e is the elementary charge, 1.6 * 10^-19 C), and the charge of the gold nucleus (q2) is +79e.
Substituting the values into the formula:
F = (9 * 10^9 N m^2 / C^2) * ((2.0e) * (79e)) / (1.7 * 10^-14 m)^2
Calculating the result:
F = (9 * 10^9) * (2.0e * 79e) / (1.7e-14)^2
F = 2.22 * 10^-10 N
Therefore, the electric force acting on the alpha particle when it is 1.7 * 10^-14 m from the gold nucleus is 2.22 * 10^-10 N.