What is the force on the charge located at x = +8.00 cm in Figure 17.40(a) given that q = 5.00 µC and a = 2.50? (The positive direction is to the right.)
To find the force on a charge located at a certain position, we can use Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is F = k * ((q1 * q2) / r^2), where F is the force, k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between them.
In this case, the charge is located at x = +8.00 cm, which means the distance between the charges will be the difference between the two positions. Given that position a = 2.50 cm, we can calculate the distance as:
r = x2 - x1 = (+8.00 cm) - (+2.50 cm) = 5.50 cm
Now, we can substitute the values into Coulomb's Law equation:
F = k * ((q1 * q2) / r^2)
F = (8.99 x 10^9 N m^2/C^2) * ((5.00 µC * 5.00 µC) / (5.50 cm)^2)
Converting the charge from microcoulombs (µC) to coulombs (C), we get:
F = (8.99 x 10^9 N m^2/C^2) * ((5.00 x 10^(-6) C * 5.00 x 10^(-6) C) / (5.50 x 10^(-2) m)^2)
Now, we can calculate the force:
F = (8.99 x 10^9 N m^2/C^2) * (2.50 x 10^(-11) C^2 / 3.025 x 10^(-4) m^2)
F = 7.43 x 10^(-3) N
Therefore, the force on the charge located at x = +8.00 cm is 7.43 x 10^(-3) N in the positive direction.