Two charged bodies exert a force of 0.245 N on each other. If they are moved so that they are one-fifth as far apart, what force is exerted?
.245/(1/5^2)= 6.125
To find the new force exerted when the charged bodies are one-fifth as far apart, we can use Coulomb's Law. According to Coulomb's Law, the force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's denote the initial distance between the charged bodies as d1 and the final distance as d2. We are told that when they are at the initial distance, the force exerted is 0.245 N. We need to find the force exerted when they are at the final distance.
According to Coulomb's Law:
F1 / F2 = (d1 / d2)^2
where F1 is the initial force and F2 is the final force.
Substituting the given values, we have:
0.245 / F2 = (d1 / d2)^2
Now, we are told that the new distance (d2) is one-fifth of the initial distance (d1). So, d2 = (1/5) * d1.
Substituting this value in the equation, we get:
0.245 / F2 = (d1 / ((1/5) * d1))^2
Simplifying further:
0.245 / F2 = (5/1)^2
0.245 / F2 = 25
Now, to find F2, we can rearrange the equation:
F2 = 0.245 / 25
Calculating this, we get:
F2 = 0.0098 N
Therefore, the force exerted when the charged bodies are one-fifth as far apart is 0.0098 N.
To find the force exerted when the distance between the charged bodies is changed, we can use Coulomb's law, which states that the force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's assume the original distance between the charged bodies is d, and the original force is F. We are given that F = 0.245 N.
When the distance is reduced to one-fifth, the new distance between the charged bodies becomes (1/5)d.
According to Coulomb's law, the force exerted is given by the equation:
F' = k * (q1 * q2) / d'^2
where F' is the new force, k is the Coulomb's constant, q1 and q2 are the charges of the bodies, and d' is the new distance between the bodies.
Since we are only comparing the change in force, we can write the equation as:
F' / F = (q1 * q2) / (d'^2 * q1 * q2)
Since the charges q1 and q2 are the same, they cancel out in the equation:
F' / F = 1 / (d'^2)
Now let's substitute the values into the equation.
F' / 0.245 N = 1 / ((1/5)d)^2
To simplify, we invert the fraction and square it:
F' / 0.245 N = (5d)^2
F' / 0.245 N = 25 * d^2
Now, we can solve for F' by multiplying both sides by 0.245 N:
F' = 0.245 N * 25 * d^2
F' = 6.125 N * d^2
Therefore, the force exerted when the charged bodies are moved one-fifth as far apart is 6.125 N times the original distance squared.
Force proportional to 1/d^2
5^2 = 25