Consider four masses arranged as described. Each is 1 cm from the center of the cross. Mass A has a charge +1µC, mass B has a charge of -1µC, and mass C has a charge of +2µC. At the center the electric field points 30 degrees from the vertical axis. What is the charge of mass D in µC?

Answer 1

To determine the charge of mass D at the center of the cross, we need to consider the electric field caused by masses A, B, and C at that point.

Let's break down the components of the electric field at the center caused by each mass:

1. Mass A (+1µC):
The electric field due to mass A at the center points directly away from mass A. Since mass A has a charge of +1µC, the electric field magnitude due to mass A at the center is given by:
E_A = k * |q_A| / r_A^2
where q_A = +1µC, r_A = 1cm = 0.01m, and k is the electrostatic constant.
Plugging in the values:
E_A = (8.99 x 10^9 Nm^2/C^2 * 1 x 10^-6 C) / (0.01m)^2 = 8.99 x 10^5 N/C

2. Mass B (-1µC):
The electric field due to mass B at the center points directly away from mass B. Since mass B has a charge of -1µC, the electric field magnitude due to mass B at the center is given by:
E_B = k * |q_B| / r_B^2
where q_B = -1µC, r_B = 1cm = 0.01m.
Plugging in the values:
E_B = (8.99 x 10^9 Nm^2/C^2 * 1 x 10^-6 C) / (0.01m)^2 = 8.99 x 10^5 N/C

3. Mass C (+2µC):
The electric field due to mass C at the center points 30 degrees from the vertical axis. The horizontal component of the electric field due to mass C at the center is:
E_Cx = E_C * cos(30°)
where E_C = k * |q_C| / r_C^2, q_C = +2µC, and r_C = 1cm = 0.01m.
Plugging in the values:
E_Cx = (8.99 x 10^9 Nm^2/C^2 * 2 x 10^-6 C) / (0.01m)^2 * cos(30°) = 7.795 x 10^5 N/C

The net electric field at the center is the vector sum of the electric field components due to each mass. Since the electric field at the center points 30 degrees from the vertical axis, we can express the vertical and horizontal components of the total electric field as:
E_vertical = E_A + E_B
E_horizontal = E_Cx

We know that the tangent of 30° is sqrt(3), so:
E_vertical = 0
E_horizontal = sqrt(3) * E

The total electric field at the center is:
E = sqrt((E_vertical)^2 + (E_horizontal)^2)
= sqrt((0)^2 + (sqrt(3) * E)^2)
= sqrt(3) * E

Plugging in the values for E:
E = sqrt(3) * 8.99 x 10^5 N/C = 1.555 x 10^6 N/C

Finally, for the net electric field at the center, we know that E = k * |q_D| / r_D^2, where q_D is the charge of mass D and r_D = 1cm = 0.01m.

Plugging in the values and solving for q_D:
1.555 x 10^6 N/C = (8.99 x 10^9 Nm^2/C^2 * |q_D|) / (0.01m)^2
|q_D| = (1.555 x 10^6 N/C * (0.01m)^2) / 8.99 x 10^9 Nm^2/C^2
|q_D| = 0.001735 C = 1.735 mC

Therefore, the charge of mass D at the center is 1.735µC.