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 the charge located at x = +8.00 cm, we need to use Coulomb's law. Coulomb's law relates the force between two charges to their magnitudes and the distance between them.
The formula for Coulomb's law is:
F = (k * |q1 * q2|) / r^2
where:
F is the magnitude of the force between the two charges,
k is Coulomb's constant (which is approximately 9.0 x 10^9 N m^2 / C^2),
q1 and q2 are the magnitudes of the charges,
and r is the distance between the charges.
In this case, q1 is the charge located at x = +8.00 cm (given as 5.00 µC), and q2 is the charge that is the source of the electrostatic field in Figure 17.40(a).
It is important to note that the distance between the charges is not given directly, but the information needed can be inferred from the figure. In this case, we can assume that the distance (r) is the difference between the position (x) of the charge we want to find the force on and the position of the other charge (a = 2.50 cm).
So, the distance between the charges is: r = x - a = (+8.00 cm) - (+2.50 cm) = 5.50 cm.
Substituting the given values into Coulomb's law, we have:
F = (9.0 x 10^9 N m^2 / C^2) * |(5.00 x 10^-6 C) * q2| / (5.50 x 10^-2 m)^2
Now we can calculate the force using the given values.