Calculating Charge Distribution in a Parallel Plate Capacitor

In summary, a small plastic ball with a mass of 6.95×10-3 kg and a charge of +0.11 mC is suspended between the plates of a capacitor. The ball is in equilibrium with the thread making a 30.0° angle with respect to the vertical. The area of each plate is 0.0152 m2. To find the magnitude of the charge on each plate, we can use the equation E=(q)/(AEo) and set up a mechanical equilibrium equation for the ball. This includes the tension force, the force of gravity, and the electric force. By using the equation G=mg and F_{el}=qE, we can solve for the magnitude of the charge
  • #1
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A small plastic ball of mass 6.95×10-3 kg and charge +0.11 mC is suspended from an insulating thread and hangs between the plates of a capacitor. 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 0.0152 m2. What is the magnitude of the charge on each plate? (the picture shows the positive plate on the left, the ball leaning 30 deg toward the negative plate, and the negative plate on the right.

With this problem, I solved for E using E=(q)/(AEo). I don't know where to go from here though. I'm confused.
 
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  • #2
Write the mechanical equilibrium of the ball
[tex]\vec{T}+\vec{G}+\vec{F_{el}}=0[/tex]
onto the horizontal and vertical direction. You now that
[tex]G=mg [/tex]
[tex]F_{el}=qE[/tex]

and [tex]E=\frac{Q}{\epsilon_0 A}[/tex]

...
 
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  • #3
oooo, i set up my vectors wrong in the beginning!
how silly of me, thank you very much for the help.
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device that stores electrical energy by accumulating positive and negative charges on two parallel conductive plates separated by an insulating material, also known as a dielectric.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by creating an electric field between the two plates when a potential difference is applied. This causes one plate to accumulate positive charges and the other plate to accumulate negative charges, creating a potential difference between the plates. The insulating material between the plates, or dielectric, helps to maintain this potential difference by reducing the strength of the electric field.

3. What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by several factors, including the distance between the plates, the surface area of the plates, and the type of dielectric material used. The capacitance increases with a decrease in plate distance and an increase in surface area. Different dielectric materials also have different permittivity values, which can affect the capacitance of the capacitor.

4. What is the equation for calculating the capacitance of a parallel plate capacitor?

The equation for calculating the capacitance of a parallel plate capacitor is C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the surface area of the plates, and d is the distance between the plates.

5. How are parallel plate capacitors used in practical applications?

Parallel plate capacitors have many practical applications, including in electronic circuits, power systems, and telecommunications. They are commonly used to store energy and regulate voltage in electronic devices, and they can also act as filters to block certain frequencies in telecommunications systems. Additionally, parallel plate capacitors are used in energy storage systems, such as batteries and supercapacitors.

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