A small m = 2.00 g plastic ball is suspended by a L = 24.0 cm long string in a uniform electric field, as shown in Figure P15.50. If the ball is in equilibrium when the string makes a 18.0° angle with the vertical as indicated, what is the net charge on the ball?

The electric field is 1x10^3 N/C

1*10^5

To find the net charge on the plastic ball, we need to consider the forces acting on it in equilibrium.

First, let's identify the forces acting on the ball. We have two forces: the weight of the ball (mg) acting downward and the electrostatic force (Fe) acting upward due to the electric field.

Since the ball is in equilibrium, the net force acting on it must be zero. Therefore, the weight of the ball must be equal and opposite to the electrostatic force.

We can start by calculating the weight of the ball using the formula: weight = mass × gravity.

Given:
Mass (m) = 2.00 g = 0.002 kg
Gravity (g) = 9.8 m/s^2

Weight (W) = 0.002 kg × 9.8 m/s^2
W ≈ 0.0196 N

Now, let's calculate the electrostatic force (Fe) using the formula: Fe = qE.

Given:
Electric field (E) = 1 × 10^3 N/C (since 1 N/C = 1 V/m)
Length of the string (L) = 24.0 cm = 0.24 m
Angle with the vertical (θ) = 18.0°

The electrostatic force (Fe) can be calculated using the equation: Fe = T × sin(θ), where T is the tension in the string.

Since the ball is in equilibrium, the tension in the string is equal to the weight of the ball. Therefore, T = W = 0.0196 N.

Fe = 0.0196 N × sin(18.0°)
Fe ≈ 0.005 N

Now that we have calculated the electrostatic force (Fe), we can use the formula Fe = qE to find the net charge (q).

0.005 N = q × 1 × 10^3 N/C

Simplifying the equation, we get:
q = 0.005 N / (1 × 10^3 N/C)
q = 5 × 10^-6 C

Therefore, the net charge on the plastic ball is 5 × 10^-6 C.