A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.162 µC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0° with respect to the vertical. The area of each plate is 0.0145 m2. What is the magnitude of the charge on each plate?
To find the magnitude of the charge on each plate, we can start by analyzing the forces acting on the plastic ball. In equilibrium, the weight of the ball is balanced by the electrostatic force due to the electric field between the plates of the capacitor.
1. First, let's calculate the weight of the plastic ball using the formula:
Weight = mass * gravitational acceleration
Weight = (7.00 * 10^-3 kg) * (9.8 m/s^2)
Weight = 6.86 * 10^-2 N
2. Next, let's find the tension in the string using the angle given. We can break down the weight into horizontal and vertical components:
Tension * cos(30°) = Weight
Tension = Weight / cos(30°)
Tension = 6.86 * 10^-2 N / 0.866
Tension = 7.92 * 10^-2 N
3. Now that we know the tension in the string, we can determine the electrostatic force on the ball. The force due to the electric field between the plates of the capacitor is given by:
Electric Force = Tension * sin(30°)
Electric Force = 7.92 * 10^-2 N * 0.5
Electric Force = 3.96 * 10^-2 N
4. Finally, let's find the magnitude of the charge on each plate. The electric field strength between the plates, E, is given by:
E = Electric Force / (charge on each plate)
Rearranging the equation, we have:
(Charge on each plate) = Electric Force / E
The area of each plate, A, is given as 0.0145 m^2, and the distance between the plates, d, is not provided. Therefore, we need to assume a separation distance between the plates in order to calculate the electric field. Let's assume a separation distance of 0.01 m.
Electric Field, E = Voltage / Distance
Voltage = Electric Field * Distance
Voltage = E * 0.01
(Charge on each plate) = Electric Force / (E * 0.01)
(Charge on each plate) = (3.96 * 10^-2 N) / (E * 0.01)
Now, we can use the equation Q = CV to relate the charge (Q) to the voltage (V) and the capacitance (C), where capacitance is given as:
C = (ε₀ * A) / d
ε₀ is the permittivity of free space and is approximately 8.85 * 10^-12 F/m.
Plugging in the values, we have:
(Charge on each plate) = (3.96 * 10^-2 N) / [(E * 0.01) * (8.85 * 10^-12 F/m * 0.0145 m^2 / 0.01 m)]
Simplifying the equation will give us the magnitude of the charge on each plate.