Calculating Potential Difference in a Suspended Parallel Plate Capacitor

In summary, a small object with a mass of 420 mg and a charge of 30.0 nC is suspended between the vertical plates of a parallel-plate capacitor with a distance of 2.00 cm between the plates. The thread of the object makes an angle of 18.0 degrees from the vertical. By using the equations PE = -q(Vb-Va) and V = kq/r, we can calculate the potential difference between the plates by finding the horizontal force on the mass and using it to determine the necessary potential difference.
  • #1
kkevinzhang
2
0

Homework Statement



A small object with a mass of 420 mg carries a charge of 30.0 nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 2.00 cm. If the thread makes an angle of 18.0° with the vertical, what is the potential difference between the plates?

Mass: 420 mg
Charge: 30.0 nC
Distance between plates: 2.00 cm
Thread of object: 18.0 degrees from vertical

Homework Equations



PE = -q(Vb-Va)
V = kq/r


The Attempt at a Solution



I just don't understand how the mass in the middle and the angle affects the problem.
 
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  • #2
welcome to pf!

hi kkevinzhang! welcome to pf! :smile:
kkevinzhang said:
I just don't understand how the mass in the middle and the angle affects the problem.

draw a https://www.physicsforums.com/library.php?do=view_item&itemid=100" to find the horizontal force on the mass …

then calculate the potential difference which would produce that force :wink:
 
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  • #3
thanks!
 

Related to Calculating Potential Difference in a Suspended Parallel Plate Capacitor

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two conducting plates that are placed parallel to each other, with a dielectric material (such as air or plastic) in between. It is used to store electrical energy by creating an electric field between the two plates.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by creating an electric field between its two plates. When a voltage is applied to the plates, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, and the capacitor is able to store energy in this field.

3. What is the formula for calculating the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates. This formula assumes that the plates are parallel and the electric field is uniform between them.

4. 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 area of the plates, the type of dielectric material used, and the voltage applied to the plates. Capacitance increases with a larger plate area, a smaller distance between the plates, and a higher dielectric constant of the material.

5. How is a parallel plate capacitor used in electronic circuits?

Parallel plate capacitors are commonly used in electronic circuits for a variety of purposes, such as filtering, energy storage, and timing. They can also be used in combination with other components to create oscillators, filters, and other types of circuits. Their ability to store electrical energy makes them an important component in many electronic devices.

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