two charges attract each other with a force of 1.5 N. what will be the force if the distance between them is reduced to one-ninth of its original value?
120N
To calculate the force between two charges, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
Let's assume the original distance between the charges is represented by 'r', and the original force is represented by 'F'.
According to the problem, we know that the original force is 1.5 N. Now, if the distance is reduced to one-ninth (1/9) of its original value, the new distance will be 1/9 of 'r'.
The formula for Coulomb's Law is:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges,
k is Coulomb's constant (k = 9 x 10^9 N.m^2/C^2),
q1 and q2 are the magnitudes of the charges, and
r is the distance between the charges.
To find the new force, we need to substitute the values into the formula. Since the charges remain the same, we can simplify the formula to:
F' = k / (r'^2)
Where:
F' is the new force,
k remains the same (9 x 10^9 N.m^2/C^2), and
r' is the new distance between the charges.
Since the distance is reduced to one-ninth of its original value, the new distance is (1/9)^2 = 1/81 of the original distance. Hence:
F' = 1.5 N * (r^2 / (1/81))^2
By simplifying this equation, we find the new force:
F' = 1.5 N * (81 / r^2)
Calculating the value further:
F' = 1.5 N * 81
F' = 121.5 N
Therefore, the force between the charges will be 121.5 N when the distance between them is reduced to one-ninth of its original value.