Two point charges are fixed on the y axis: a negative point charge q1 = -22 µC at y1 = +0.24 m and a positive point charge q2 at y2 = +0.37 m. A third point charge q = +7.7 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 24 N and points in the +y direction. Determine the magnitude of q2.
. 18.0 × 10−8 C,
To determine the magnitude of q2, we can start by analyzing the forces acting on charge q caused by charges q1 and q2.
Let's assume the distance between charge q and charge q1 is r1, and the distance between charge q and charge q2 is r2.
According to Coulomb's Law, the electrostatic force between two point charges is given by F = k * |q1 * q2| / r^2, where k is the electrostatic constant (k ≈ 9 × 10^9 N m^2 C^-2).
For charge q and charge q1, the force can be expressed as:
F1 = k * |q * q1| / r1^2
Similarly, for charge q and charge q2, the force can be expressed as:
F2 = k * |q * q2| / r2^2
Given that the net electrostatic force exerted on charge q (F_net) has a magnitude of 24 N and points in the +y direction, we can split it into its y-components:
F_net = F1_y + F2_y
Now, let's analyze the forces in terms of their y-components:
F1_y = F1 * sin(θ1)
F2_y = F2 * sin(θ2)
Since F_net points in the +y direction, F_net_y = 24 N.
Since these forces are in different directions, we'll take the magnitudes into account:
F1_y = |F1| * sin(θ1)
F2_y = |F2| * sin(θ2)
From the given information, we can identify that charge q1 is negative, and charge q2 is positive. The net force points in the +y direction, which means the force due to charge q1 (F1) is in the -y direction, and the force due to charge q2 (F2) is in the +y direction.
Now, we have:
F1 = -F1_y
F2 = F2_y
Substituting these values into the equation for F_net_y:
F_net_y = -F1_y + F2_y
24 N = -F1_y + F2_y
We also know the values of the charges q1 and q2:
q1 = -22 µC
q2 = ?
To calculate the electric forces F1 and F2, we need to find the distances r1 and r2. We can use the given information for the y-positions of the charges:
y1 = +0.24 m
y2 = +0.37 m
Since the charges are aligned on the y-axis and the charge q is located at the origin (0, 0), the distances r1 and r2 are simply the y-coordinates of the charges:
r1 = y1 = +0.24 m
r2 = y2 = +0.37 m
Now, we can calculate the forces F1 and F2 using Coulomb's Law:
F1 = k * |q * q1| / r1^2
F2 = k * |q * q2| / r2^2
Plugging in the values:
F1 = k * |-7.7 µC * -22 µC| / (0.24 m)^2
F2 = k * |-7.7 µC * q2| / (0.37 m)^2
Solving for F_net_y:
24 N = -F1_y + F2_y
24 N = -F1 * sin(θ1) + F2 * sin(θ2)
Substituting the forces:
24 N = F1 * sin(θ1) + F2 * sin(θ2)
Plugging in F1 and F2:
24 N = (k * |-7.7 µC * -22 µC| / (0.24 m)^2) * sin(θ1) + (k * |-7.7 µC * q2| / (0.37 m)^2) * sin(θ2)
We have all the values except q2. We can calculate q2 by rearranging the equation and solving for q2:
q2 = ((24 N / sin(θ2)) - (k * |-7.7 µC * -22 µC| / (0.24 m)^2) * sin(θ1)) * (0.37 m)^2 / (k * |-7.7 µC|)
Calculating this expression will give us the magnitude of q2.