A small plastic ball with a mass of 6.50x10^-3 kg and with a charge of +0.150x10^-6 C is suspended from an insulating thread and hangs between the plates of a capacitor. The ball is in equilibrium, whit the tread making an angle of 30 degrees with respect to the vertical. The area of each plate is 0.0150 m^2. What is the magnitude of the charge of each plate?
To find the magnitude of the charge on each plate, we can use the information given and apply the concept of equilibrium.
In equilibrium, the gravitational force acting on the ball should be balanced by the electrostatic force between the ball and the plates of the capacitor.
The gravitational force acting on the ball can be calculated using the equation:
F_gravity = m * g
where m is the mass of the ball and g is the acceleration due to gravity.
F_gravity = (6.50x10^-3 kg) * (9.8 m/s^2)
F_gravity = 6.37x10^-2 N
The electrostatic force between the ball and the plates of the capacitor can be calculated using the equation:
F_electrostatic = q * E
where q is the charge on the ball and E is the electric field strength between the plates.
The electric field strength between the plates can be calculated using the equation:
E = V / d
where V is the potential difference between the plates and d is the distance between them.
Since the ball is in equilibrium, the electrostatic force should be equal to the gravitational force:
q * E = m * g
We can rearrange this equation to solve for the charge on the ball:
q = (m * g) / E
Now, we need to find the potential difference between the plates. We can use the equation:
V = Ed
where E is the electric field strength and d is the distance between the plates.
Given that the area of each plate is 0.0150 m^2, we can calculate the distance between the plates using the equation:
d = A / (2 * Δ)
where A is the area of each plate and Δ is the separation between them (which is unknown).
So, we have:
d = (0.0150 m^2) / (2 * Δ)
Now, we substitute the value of d in the equation V = Ed:
V = E * (0.0150 m^2) / (2 * Δ)
V = E * (0.0150 m^2) / (2 * d)
Since the electric field strength is the same throughout the region between the plates, we can simplify this to:
V = E * (0.0150 m^2 / d)
Now, let's substitute the value of V in the equation q = (m * g) / E:
q = (m * g) / (E * (0.0150 m^2 / d))
Given that the ball has a charge of +0.150x10^-6 C, we can solve for the magnitude of the charge on each plate:
+0.150x10^-6 C = (6.50x10^-3 kg * 9.8 m/s^2) / (E * (0.0150 m^2 / d))
Simplifying further:
0.150x10^-6 C = (6.37x10^-2 N) / (E * (0.0150 m^2 / d))
Now, we need to substitute the value of E using the equation E = V / d:
0.150x10^-6 C = (6.37x10^-2 N) / ((V * d) / d * (0.0150 m^2 / d))
Simplifying further:
0.150x10^-6 C = (6.37x10^-2 N) / (0.0150 m^2 / d)
To find the magnitude of the charge on each plate, we need to solve for d. However, it seems that the value of d is missing from the given information.
To find the magnitude of the charge of each plate, we can use the concept of equilibrium in an electric field.
Here's how we can approach the problem step by step:
Step 1: Identify the forces acting on the charged ball.
The two main forces acting on the ball are the gravitational force (mg) and the electrostatic force due to the electric field between the plates (Eq). Since the ball is in equilibrium, these two forces must be equal and opposite.
Step 2: Break down the forces into their components.
The gravitational force can be broken down into two components: one parallel to the thread (mg*sinθ) and one perpendicular to the thread (mg*cosθ).
Step 3: Equate the forces.
To find the magnitude of the charge on each plate, we need to equate the gravitational and electrostatic forces acting on the ball:
mg*sinθ = Eq
Step 4: Substitute the given values.
We are given:
Mass of the ball (m) = 6.50 × 10^-3 kg
Charge on the ball (q) = +0.150 × 10^-6 C
Angle (θ) = 30 degrees
Area of each plate (A) = 0.0150 m^2
Substituting these values into the equation, we get:
(6.50 × 10^-3 kg) * (9.8 m/s^2) * sin(30 degrees) = (E) * (0.150 × 10^-6 C)
Step 5: Solve for the electric field (E).
Rearrange the equation to solve for E:
E = [(6.50 × 10^-3 kg) * (9.8 m/s^2) * sin(30 degrees)] / (0.150 × 10^-6 C)
Step 6: Calculate the electric field (E).
Calculating the right-hand side of the equation, we get:
E ≈ 2.97 × 10^4 N/C
Step 7: Calculate the magnitude of the charge on each plate.
Since the electric field between the plates of a capacitor is given by E = Q / A, where Q is the magnitude of the charge on each plate and A is the area of each plate, we can rearrange the equation to solve for Q:
Q = E * A
Substitute the given values:
Q = (2.97 × 10^4 N/C) * (0.0150 m^2)
Step 8: Calculate the magnitude of the charge on each plate.
Calculating the right-hand side of the equation, we get:
Q ≈ 4.46 × 10^-3 C
Therefore, the magnitude of the charge on each plate is approximately 4.46 × 10^-3 C.