An electron that is a certain distance from a proton is acted on by an electrical force.
a) If the electron were moved twice the distance away from the proton, would the electrical force be:
1/4
because of
k Q1 Q2 / r^2
2^2 is 4
Try to answer these yourself first please. The other one you posted uses the same force logic.
To answer this question, we can use Coulomb's Law, which states that the electrical force between two charged particles is inversely proportional to the square of the distance between them.
Coulomb's Law formula:
F = k * (q1 * q2) / r^2
Where:
F = electrical force
k = Coulomb's constant (approximately 9 x 10^9 N * m^2 / C^2)
q1, q2 = charges of the electron and proton respectively
r = distance between the electron and the proton
Now, let's consider the scenario in question: If the electron were moved twice the distance away from the proton.
Let the original distance between the electron and proton be 'r'. The electrical force is given by:
F1 = k * (q1 * q2) / r^2
If the electron is moved twice the distance away, the new distance will be '2r'. The electrical force at this new distance is given by:
F2 = k * (q1 * q2) / (2r)^2 = k * (q1 * q2) / 4r^2
To compare F1 and F2, we can divide the second equation by the first equation:
F2 / F1 = [k * (q1 * q2) / (4r^2)] / [k * (q1 * q2) / r^2]
= (k * (q1 * q2) / 4r^2) * (r^2 / (k * (q1 * q2)))
= 1 / 4
Therefore, the electrical force when the electron is moved twice the distance away from the proton is one-fourth (1/4) of the original electrical force.
In conclusion, if the electron is moved twice the distance away from the proton, the electrical force between them would be reduced to one-fourth of its original value.