Two charges that are 2 meters apart repel each other with a force of 2 x 10-5 newton. If the distance between the charges is
decreased to 1 meter, the force of repulsion will be
A) 1 x 10-5 N
B) 4 x 10-5 N
C) 8 x 10-5 N
D) 5 x 10-6 N
8x10^-5N
IT'S DEFINITELY "C" BECAUSE I GOT IT RIGHT ON A TEST.
8x10^-5N Is the right answer
is C trust me
its C! promise <3
it was C thanks Shakina
Yea it's def c
It is c
To determine the force of repulsion when the distance between the charges is decreased to 1 meter, we can use Coulomb's law, which states that the 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.
Let's calculate the initial charge and force of repulsion using the given information:
Given:
Distance between the charges (r1) = 2 m
Force of repulsion (F1) = 2 x 10^-5 N
Now, let's calculate the final force of repulsion when the distance between the charges is decreased to 1 meter:
Distance between the charges (r2) = 1 m
Using Coulomb's law:
F ∝ q1 * q2 / r^2, where F is the force of repulsion, q1 and q2 are the charges, and r is the distance between the charges.
Since we want to compare the force when the distance is decreased to 1 meter, we can write:
F2 / F1 = (q1 * q2 / r2^2) / (q1 * q2 / r1^2)
F2 / F1 = (q1 * q2 * r1^2) / (q1 * q2 * r2^2)
F2 / F1 = (r1^2 / r2^2)
Now substituting the given values:
F2 / F1 = (2^2 / 1^2)
F2 / F1 = 4 / 1
F2 / F1 = 4
So, the force of repulsion when the distance between the charges is decreased to 1 meter (F2) will be 4 times the initial force of repulsion (F1).
The initial force of repulsion (F1) is given as 2 x 10^-5 N, we can multiply it by 4 to find F2:
F2 = 4 * (2 x 10^-5 N)
F2 = 8 x 10^-5 N
Therefore, the force of repulsion when the distance between the charges is decreased to 1 meter will be 8 x 10^-5 N.
Thus, the correct answer is C) 8 x 10^-5 N.