Electric field in plate capacitors

In summary, two questions were discussed in this conversation. The first question involved a small plastic ball suspended between the plates of a capacitor, and the goal was to find the magnitude of the charge on each plate. The second question involved two point charges fixed at opposite corners of a square, with a test charge released from rest at one corner and the goal being to determine its speed when it reaches the center. Coulomb's law and energy conservation were used to solve the two problems.
  • #1
alreadytaken
2
0

Homework Statement



A small plastic ball of mass 5.54 x 10-3 kg and charge +0.174C 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 30.0o with respect to the vertical. The area of each plate is 0.02030 m^2. What is the magnitude of the charge on each plate?



ch18p_43.gif


Homework Equations



Electric Field = Force/test charge Electric field in plate capacitor is sigma/permitivity of space

The Attempt at a Solution



So...
Tsin30 = mgcos30 becuz in equilibrium
T = 0.0940 which the Force then divide that by test charge gives you the electric field in that area (plate capacitor).

that quotient = charge on plate/epsilon naught * (area of plate) = my answer but its not right. Acutal answer is 3.24 x 10^-8 C
 
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  • #2
I know I am being greedy but there is another problem I need solved but will have to wait all weekend. I have a poor physics background and I am taking the mcat so any input would be very gracious.

Question: Two identical point charges (q = +9.10 x 10-6 C) are fixed at opposite corners of a square whose sides have a length of 0.460 m. A test charge (q0 = -5.30 x 10-8 C), with a mass of 7.10 x 10-8 kg, is released from rest at one of the corners of the square. Determine the speed of the test charge when it reaches the center of the square

EAT_12350644631160_9437698556103365.gif


i got an answer of -612 m/s?
 
  • #3
are you taking into account the force of 1 plate on the ball or both? one plate pushes the other pullsthe second question you should draw a force diagram and use coulomb's law to find force and then acceleration
 
  • #4
For 1st
Draw FBD including firce due to both plates, weight, and tension,

Find X and Y component of tension and use the info that ball is in equilibrium
 
  • #5
second,
use energy conservation
 

1. What is an electric field in a plate capacitor?

The electric field in a plate capacitor is a region in space where an electric charge experiences a force. In a plate capacitor, this field is created by the separation of positive and negative charges on two parallel plates.

2. How is the electric field in a plate capacitor calculated?

The electric field in a plate capacitor can be calculated by dividing the voltage between the plates by the distance between them. This is known as the electric field strength and is measured in volts per meter (V/m).

3. How does the distance between the plates affect the electric field in a plate capacitor?

The electric field in a plate capacitor is directly proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field strength decreases.

4. What is the role of dielectric material in a plate capacitor's electric field?

A dielectric material is an insulating material that is placed between the plates of a capacitor. It increases the capacitance of the capacitor and decreases the electric field strength between the plates.

5. Can the electric field in a plate capacitor be manipulated?

Yes, the electric field in a plate capacitor can be manipulated by changing the voltage, distance between plates, or the type of dielectric material used. This allows for control over the strength and direction of the electric field, making plate capacitors useful in various applications such as in electronic circuits and energy storage.

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