When point charges q1 = +5.4 ìC and q2 = +8.3 ìC are brought near each other, each experiences a repulsive force of magnitude 0.69 N. Determine the distance between the charges.
i used the eqution F=k(q1)(q2)/r^2 . i got 7.64E5 but it is not right .can you please help
Did you convert the micro-coloumbs to coloumbs? The equation worked fine for me.
To find the distance between the charges, we can rearrange the formula for the electrostatic force between two point charges as follows:
F = k(q1)(q2) / r^2
Where:
- F is the force between the charges
- k is the electrostatic constant (k = 9 × 10^9 N m^2/C^2)
- q1 and q2 are the magnitudes of the charges
- r is the distance between the charges
In this case, we know that the force (F) is 0.69 N, q1 is +5.4 μC, and q2 is +8.3 μC. We need to solve for r.
0.69 N = (9 × 10^9 N m^2/C^2) * (5.4 μC) * (8.3 μC) / r^2
First, let's convert the charges to coulombs by dividing the given values by 10^6:
0.69 N = (9 × 10^9 N m^2/C^2) * (5.4 × 10^-6 C) * (8.3 × 10^-6 C) / r^2
Simplifying further:
0.69 N = (9 × 10^9 × 5.4 × 8.3 × 10^-12) / r^2
0.69 N = (37.674 × 10^-3) / r^2
Now, let's solve for r^2:
r^2 = (37.674 × 10^-3) / 0.69
r^2 ≈ 54.65 × 10^-3
Taking the square root of both sides to find r:
r ≈ √(54.65 × 10^-3)
r ≈ 0.234 m
Therefore, the distance between the charges is approximately 0.234 meters.