Parallel-Plate Capacitor

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In summary: Oops, my bad, I misread the question, I though the plates were at an angle :frown: . You're right, and its time for me to take a break...:rolleyes:
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Homework Statement


A small object with a mass of 350mg carries a charge of 30nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separacted by 4 cm. If the thread makes an angle of 15 degrees with the vertical, what is the potential difference between the plates?


The Attempt at a Solution



First, I drew a diagram. Since the object has a + charge, it would be more closer to the negative plate.

[tex]\Delta V=\frac{U_{E}}{q}[/tex]

[tex]U_{E}=qEd[/tex] , [tex]E=\frac{kq}{r^2}[/tex]

Now that I have all the required info to solve for the electrical potential energy, I solved it and put it into the equation: [tex]\Delta V=\frac{U_{E}}{q}[/tex]

For the electrical potential energy I got: 2.023*10^-4J

For the answer (the potential difference) I got: 6742.5V

Does my method seem correct? This is an even problem so I have no way of checking my answer for accuracy. Thanks in adv. :smile:
 
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  • #2
Do you use the mass of the object in your solution...:wink:
 
  • #3
Hootenanny said:
Do you use the mass of the object in your solution...:wink:

xD I knew that I left something out.. So does that mean my solution doesn't work? I don't know how to use mass in this equation. Do I have to use forces or something?? F=ma?
 
  • #4
AznBoi said:
xD I knew that I left something out.. So does that mean my solution doesn't work? I don't know how to use mass in this equation. Do I have to use forces or something?? F=ma?
Indeed, if the mass is suspended (i.e. stationary) what can you say about the weight of the mass in relation to the force exerted on it by the electric field?
 
  • #5
Hootenanny said:
Indeed, if the mass is suspended (i.e. stationary) what can you say about the weight of the mass in relation to the force exerted on it by the electric field?

Well you can use the weight to find the y component of tension and then solve for the x component of tension. Then you use F=ma with the electric force exerted opposite of the x component of tension?

So will it be:
[tex]\Sigma F_{x}=0[/tex]

[tex]-T_{x}+F_{e}=0[/tex]

[tex]F_{e}=T_{x}[/tex]

Then you use this equation:
[tex]\Delta V=\frac{U_{E}}{q}[/tex]

Instead, you solve the electrical potential energy with the electric force?

[tex]U_{E}=F_{e}*d[/tex]
 
  • #6
Sounds spot on to me, but don't forget the electric force will also contribute towards the forces in the y direction :approve:
 
  • #7
Hootenanny said:
Sounds spot on to me, but don't forget the electric force will also contribute towards the forces in the y direction :approve:

Alrighty, thanks for your help! =] But I don't get how the electric force contributes to the forces in the y direction. Aren't the plates parallel and vertical? So wouldn't the electric force be horizontal, therefore only contribute to the x direction? I 'm probably wrong, but that's how I invision it to be.
 
  • #8
AznBoi said:
Alrighty, thanks for your help! =] But I don't get how the electric force contributes to the forces in the y direction. Aren't the plates parallel and vertical? So wouldn't the electric force be horizontal, therefore only contribute to the x direction? I 'm probably wrong, but that's how I invision it to be.
Oops, my bad, I misread the question, I though the plates were at an angle :frown: . You're right, and its time for me to take a break...:rolleyes:
 

What is a parallel-plate capacitor?

A parallel-plate capacitor is a type of electronic component that stores electrical energy. It consists of two parallel conductive plates separated by a dielectric material. When a voltage is applied to the plates, an electric field is created between them, allowing the capacitor to store energy.

What factors affect the capacitance of a parallel-plate capacitor?

The capacitance of a parallel-plate capacitor is affected by three main factors: the size of the plates, the distance between the plates, and the type of dielectric material used. Generally, the larger the plates and the closer they are together, the higher the capacitance. Different materials also have different dielectric constants, which can affect the capacitance.

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

The formula for calculating the capacitance of a parallel-plate capacitor is C = ε₀A/d, where C is the capacitance, ε₀ is the permittivity of free space, 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.

What is the maximum capacitance that a parallel-plate capacitor can have?

The maximum capacitance of a parallel-plate capacitor is theoretically infinite, as long as the plates are infinitely large and infinitely close together. However, in practical applications, the maximum capacitance is limited by the breakdown voltage of the dielectric material used, which is the maximum voltage the capacitor can withstand before the dielectric breaks down.

What are some common uses of parallel-plate capacitors?

Parallel-plate capacitors have a wide range of uses in electronic circuits. They are commonly used in power supplies to filter out unwanted noise and provide stable voltage, in radio-frequency circuits for tuning and filtering, and in electronic filters for audio and video signals. They are also used in sensor applications, such as capacitive touch screens and accelerometers.

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