Charged ball hanging in equilibrium between two parallel plate capacitors.

In summary, an experiment was conducted to test the validity of electrostatic principles using a small ball with a charge of ±1.50 μC suspended between two parallel plate capacitors. The signs of the charges on the plates and ball were determined, and a free body diagram was drawn for the ball. The magnitude of the electric field produced by the capacitor was found to be 3.68 x 10-2N, and the potential difference between the plates was calculated to be 1.00 x 10-4m. The surface area of each plate was determined to be 10.0 C, and the amount of flux passing through one plate was calculated. If the experiment were immersed in non-conductive oil, the
  • #1
bitterbilly
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1. An experiment is run to determine the validity of numerous electrostatic principles. A small ball of mass 6.50 x 10-3 kg with a charge of ±1.50 μC is suspended from a non-conductive wire and hangs between two parallel plate capacitors. An angle of 30.0° is measured between the wire and the vertical while the ball is in equilibrium.
a. Determine the sign of the charge on each of the plates and the small ball.
b. Draw a free body diagram for the small ball.
c. Determine the magnitude of the electric field produced by the parallel plate capacitor.
d. Solve for the potential difference between the two plates that are separated by a distance of 1.00 x 10-4m.
e. A charge of 10.0 C is stored on each plat. Determine the surface area of each plate.
f. Calculate the amount of flux that passes through one plate.
g. Discuss the change in the angle of θ if the same experiment were immersed in non-conductive oil. (Assume the dielectric constant for the oil is greater than that of air.)


I am having a lot of trouble trying to figure out what to do.



So far, for (a) I have the signs as + ---------> -
+ ---------> - etc.
For (b) I have weight going down, the Electric Force, FE going right, and Tension going North/West at 30°.

For (c) I set Fy = W, and so Fy = 6.37 x 10-2N. Then using trig I found that Fx = 3.68 x 10-2N, which is then equal to the Force.

This is where I'm stuck, I don't really know how to find the value for the E field. Any help would be much appreciated.
 
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  • #2
I agree with your Fy and Fx. Fx is the horizontal electric force on the 1.5μC charge.
The equation for the force on a charge, q, in an electric field, E is F = q x E.
You should be able to find the Electric field strength E.
Field strength is measured in Volts per metre so if you know the separation of the plates you should be able to find the voltage across them.
BUT... part (d) in your question states that the plates are 1 x 10^-4 m apart. This is only 1/10 of a mm... are you sure about this?
 

1. How can a charged ball hang in equilibrium between two parallel plate capacitors?

This phenomenon occurs due to the balance of electrical forces acting on the charged ball. The electric field between the two parallel plate capacitors exerts a force on the ball, causing it to move towards one of the plates. However, the electric charge on the ball also creates an electric field that repels it from the plate. When these two forces are equal and opposite, the ball remains suspended in equilibrium.

2. What factors affect the equilibrium position of the charged ball?

The equilibrium position of the charged ball is influenced by the distance between the two plates, the charge on the ball, and the capacitance of the parallel plate capacitors. As these factors change, the balance of forces and the position of the ball will also change.

3. Can the equilibrium position of the charged ball be adjusted?

Yes, the equilibrium position of the charged ball can be adjusted by changing the voltage applied to the parallel plate capacitors or by altering the charge on the ball. This will cause a change in the electric field and force acting on the ball, resulting in a new equilibrium position.

4. What happens if the charge on the ball is increased?

If the charge on the ball is increased, the electric field created by the ball will also increase. This will cause the repulsive force from the plate to increase, and the equilibrium position of the ball will shift towards the other plate.

5. How is the equilibrium position of the charged ball affected by the shape of the parallel plate capacitors?

The shape of the parallel plate capacitors does not directly affect the equilibrium position of the charged ball. However, the distance between the plates and the capacitance of the capacitors, which can be influenced by their shape, do play a role in determining the equilibrium position of the ball.

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