If you increase the distance between two identical point charges so that the new distance is five times the original distance, what happens to the force between them?
I know it is suppose to be divided by 25 but why?
Coulomb's Law
F1=k•q1•q2/r²
F2=k•q1•q2/R²
If R=5r, then F2=F1/25
It's called an inverse square law.
Gravity of a point mass behaves the same way.
When you increase the distance between two identical point charges, the force between them decreases. This relationship is governed by Coulomb's law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, Coulomb's law is stated as:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges,
k is the proportionality constant (Coulomb's constant),
q1 and q2 are the magnitudes of the charges, and
r is the distance between the charges.
If the new distance is five times the original distance (r_new = 5 * r_original), we can see the effect on the force by plugging in the new distance into Coulomb's law:
F_new = k * (q1 * q2) / (r_new)^2
Since r_new = 5 * r_original, we have:
F_new = k * (q1 * q2) / (5 * r_original)^2
Simplifying this equation:
F_new = k * (q1 * q2) / (25 * r_original^2)
This shows that when the distance is increased by a factor of 5, the force between the charges decreases by a factor of 25 (5^2), as the distance is squared in the denominator of Coulomb's law.
To understand why the force between two point charges decreases by a factor of 25 when the distance between them is increased by a factor of 5, we need to consider Coulomb's Law.
Coulomb's Law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It can be mathematically expressed as:
F = k * (q1 * q2) / r^2
Where:
- F is the force between the charges
- k is Coulomb's constant, a proportionality constant equal to approximately 9 × 10^9 N•m²/C²
- q1 and q2 are the magnitudes of the two charges
- r is the distance between the charges
In the given scenario, the point charges are identical, which means they have the same magnitude, denoted as q (q1 = q2 = q). The original distance between them is denoted as r, and the new distance is 5 times the original distance (r_new = 5r).
To compare the forces, we can set up the following ratio:
F_new / F = (k * (q * q) / (r_new^2)) / (k * (q * q) / r^2)
Now, let's simplify this expression:
F_new / F = ((k * (q * q)) / (r_new^2)) * (r^2 / (k * (q * q)))
Notice that the q's and k's in the numerator and denominator cancel out:
F_new / F = (1 / (r_new^2)) * (r^2 / 1) = r^2 / (r_new^2) = r^2 / (5r)^2 = r^2 / (25r^2)
Finally, we can simplify further by canceling out the common factor of r^2:
F_new / F = 1 / 25
This shows that the force between the charges, F_new, is 1/25th of the original force, F.
Therefore, the force between two identical point charges decreases by a factor of 25 when the distance between them is increased by a factor of 5.