EARLIER SOMEONE POSTED:
"Two like-charged balloons, placed at a distance of .5 meters, experience a repulsive force of .32 N. What is the force if the distance between the two balloons is doubled?
If you use Coulomb's Law, the equation used to find .32 N was Force = the constant of K multiplied by the product of the charges divided by the square of the distance. Therefore, I think to solve this, you need to double r, which would mean that force is one-fourth of what it originally was. Following that, I think you divide .32 N by 4 and the answer would be .08 N =F."
I WAS WONDERING WHY IT WOULDN'T BE 4 multiplied by the original force of .32 N SO THAT THE ANSWER WOULD BE 1.28 N.
Can someone PLEASE explain why this is?
Sure! Let me explain why the force would not be 4 times the original force of 0.32 N when the distance between the two balloons is doubled.
According to Coulomb's Law, the force between two charges is inversely proportional to the square of the distance between them. Mathematically, this can be expressed as:
Force = (K * q1 * q2) / r^2
Where:
- Force is the electrostatic force between the charges
- K is the electrostatic constant
- q1 and q2 are the charges of the two objects
- r is the distance between the charges
Now, let's consider the given situation with two like-charged balloons.
Initially, the distance (r) between the balloons is 0.5 meters, and the force is 0.32 N. Let me call this initial force F1.
Now, when the distance between the balloons is doubled to 1 meter, let's call this new force F2. We want to find the value of F2.
To use Coulomb's Law, we can compare the ratios of the distances:
(r2 / r1) = (1 m / 0.5 m) = 2
Since distance is in the denominator of Coulomb's Law, when the distance doubles, the force is actually divided by 2^2 = 4, not multiplied by 2 or 4.
So, to find the new force F2, we can divide the initial force F1 by 4:
F2 = F1 / 4 = 0.32 N / 4 = 0.08 N
Therefore, the correct answer is 0.08 N, as you initially mentioned in your explanation. The force is one-fourth of the original force, not four times.