A small plastic ball with a mass of 6.80 10-3 kg and with a charge of +0.161 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 0.0155 m2. What is the magnitude of the charge on each plate?
m=6.8•10⁻³ kg, q=0.161•10⁻⁶C, A=0.0155 m², α=30°, Q=?
Let’s find the projections of the forces acting on the plastic ball:
x-projections: Tsinα =F ……. (1)
y-projections: Tcosα = mg ……(2)
Divide (1) by (2)
Tsinα /Tcosα=F/mg,
tanα =F/mg…………(3)
F=qE=qσ/ε₀=q•Q/A ε₀ …(4)
Substitute (4) in (3)
tanα = qQ/A• ε₀•m•g,
Q= A•ε₀•mg• tanα/q=
=0.0155•8.85•10⁻¹²•6.8•10⁻³•9.8•0.577/0.161•10⁻⁶=
= 3.28•10⁻⁸ C
To find the magnitude of the charge on each plate, we can start by analyzing the forces acting on the suspended plastic ball.
When the ball is in equilibrium, the gravitational force pulling it downwards is balanced by the electrical force due to the electric field between the plates of the capacitor.
Let's break down the forces involved:
1. Gravitational force (Fg):
The gravitational force acting on the ball is given by the equation:
Fg = mass * acceleration due to gravity
In this case, the mass of the ball is 6.80 * 10^-3 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore, we can calculate the gravitational force acting on the ball.
2. Electrical force (Fe):
The electrical force acting on the ball is given by the equation:
Fe = charge on the ball * electric field strength
The electric field strength between the plates of the capacitor can be calculated by dividing the voltage (V) applied to the capacitor by the distance (d) between the plates:
E = V / d
In this case, the electric field strength (E) can be found using the given voltage (V) and the distance (d) between the plates.
The charge on each plate can be calculated as follows:
3. Force equilibrium:
For the ball to be in equilibrium, the gravitational force (Fg) and the electrical force (Fe) must be equal and opposite. Therefore, we can equate the equations for Fg and Fe to find the charge on the plates.
Now that we have broken down the problem into smaller parts, we can proceed to determine the magnitude of the charge on each plate.