- #1

giladsof

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- 0

## Homework Statement

## Homework Equations

U = 0.5CV^2

## The Attempt at a Solution

I'm really lost on this one...?

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In summary: So, you're actually doing work to move the charges, and the more charges you move, the more work you do. The equation for the force on a charge in an Electric Field is F = qE. The equation for the work done by a force through a distance is W = qD. Each charge has to be treated separately in this equation- so there are two equations.

- #1

giladsof

- 13

- 0

U = 0.5CV^2

I'm really lost on this one...?

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- #2

zhermes

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Yes, you do seem to be fairly lost on this one.

Is that the only question you had?

Is that the only question you had?

- #3

giladsof

- 13

- 0

But in this problem I don't seem to manage to do so... more help and less sarcastic remarks will be most appreciative

- #4

berkeman

Mentor

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- 21,814

giladsof said:## Homework Statement

## Homework Equations

U = 0.5CV^2

## The Attempt at a Solution

I'm really lost on this one...?

Do you see intuitively whether it will require energy to be put into the system, or give energy back, to switch the position of the plates? Why?

And to calculate it, I think I would just calculate the energy required to move the charges those distances in that Electric Field. What is the equation for the force on a charge in an Electric Field? What is the equation for the work done by a force through a distance? Don't forget to account for each plate's charge separately (so two energy calculations).

zhermes said:Yes, you do seem to be fairly lost on this one.

Is that the only question you had?

I believe he is just gently reminding you of the PF Homework Help rules, which require you to show an attempt at a solution.

- #5

giladsof

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One small question remains- I don't understand why is it so intuitive that the capacitor will resist this change?

- #6

berkeman

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giladsof said:

One small question remains- I don't understand why is it so intuitive that the capacitor will resist this change?

Because you're having to move charges against the forces from the E-field. You have to do work to move the charges against those forces. If you were going the other way, well, that's the way the charges would go anyway if they were not constrained to be held on the plates (if they were free charges).

A parallel-plate capacitor with flipping charge is a type of capacitor that consists of two parallel plates separated by a dielectric material. It is called "flipping" because the charge on each plate can be flipped or reversed by changing the polarity of the voltage source.

When a voltage is applied to the capacitor, electrons from the negative plate are attracted to the positive plate, creating a potential difference between the plates. This creates an electric field between the plates, which stores energy in the form of electric potential energy. The flipping charge occurs when the polarity of the voltage source is changed, causing the electrons to switch plates, thus reversing the charge on each plate.

The capacitance of a parallel-plate capacitor with flipping charge can be calculated using the equation C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

Parallel-plate capacitors with flipping charge are commonly used in electronic circuits for energy storage, filtering, and signal processing. They are also used in various technologies such as touch screens, flash memory, and sensors.

The capacitance and flipping charge of a parallel-plate capacitor can be affected by factors such as the distance between the plates, the area of the plates, and the dielectric material used. Additionally, the voltage and frequency of the applied signal can also impact the capacitance and flipping charge of the capacitor.

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