# A Question Related to Electrostatic Energy and Capacitors

• Air_Gear
In summary, the given problem involves three charges, Q, Qx, and Qy, where Qx is twice the magnitude of Q and Qy is unknown. The work required to bring in Qy is twice that of bringing in Qx. Using the formula for work, (W = k * Qx * Qy / R), it is determined that Qy is equal to 2Q. The vectors involved in this problem do not affect the solution.
Air_Gear

## Homework Statement

A charge Q is at the origin. A second charge, Qx = 2Q, is brought to the point x = a and y = 0. A third charge Qy is brought to the point x = 0, y = a. If it takes twice as much work to bring in Qy as it did Qx, what is Qy in terms of Q?

## Homework Equations

My main problem in this (Besides an incompetent professor ) is connecting the relationships of Qy, Qx, and Q, specifically Qy and Q. Also, I believe my formula for work to bring in a charge

(W = (k * Qx * Qy)/ R)

Might be wrong.

## The Attempt at a Solution

W = work to bring in charge.

Wy = 2*Wx

Wx = (k * Q * Qx)/a => Wx = (k * 3Q)/a

Wy = (k * Q * Qy)/a

(k * Q * Qy)/a = 2 * (k * 3Q)/a => (k * Q * Qy)/a = (k * 6Q)/a

And here's where I start to have problems; constant 'k' and 'a' get canceled out, fine and dandy. However, the 'Q' also gets canceled out, which is bad as that's what I want my final answer in terms of.

Also, am I over thinking this, or do the vectors of Qy and Qx not allow me to make this connection?

Thanks to anyone can help, and I hope to have a pleasant stay here at Physics Help

Welcome to pf :)

Wx = (k * Q * Qx)/a => Wx = (k * 3Q)/a
Check the math.
Also, from the question it is not clear whether Q2 was brought with both Q and Q1 in place or Q alone.

Thank you!

The problem didn't specify, but from the wording I would imagine that Qy was brought in by Q alone.

Is the math correct now?

Wx = (k * Q * Qx)/a => Wx = (k * 2Q^2)/a

Air_Gear said:
Is the math correct now?

Wx = (k * Q * Qx)/a => Wx = (k * 2Q^2)/a

Looks good to me!

The problem didn't specify, but from the wording I would imagine that Qy was brought in by Q alone.
Funny, I would have thought the other, from the wording of the question. Doesn't matter, as long as you know how to solve for both ;)

## 1. What is electrostatic energy and how is it related to capacitors?

Electrostatic energy is the potential energy stored in a system of electric charges due to their positions. Capacitors, which are electronic components that store electric charge, utilize electrostatic energy to function. As charges are stored on the plates of a capacitor, electrostatic energy is stored in the form of an electric field between the plates.

## 2. How is the amount of electrostatic energy in a capacitor determined?

The amount of electrostatic energy stored in a capacitor is determined by the capacitance of the capacitor and the voltage applied to it. The higher the capacitance and voltage, the more electrostatic energy is stored in the capacitor.

## 3. Can the electrostatic energy in a capacitor be converted into another form of energy?

Yes, the electrostatic energy in a capacitor can be converted into other forms of energy such as heat, light, or kinetic energy. This is often seen in electronic devices, where the stored electrostatic energy is used to power the device.

## 4. How does the dielectric material in a capacitor affect its electrostatic energy?

The dielectric material in a capacitor affects its electrostatic energy by increasing the capacitance of the capacitor, thus allowing for more electrostatic energy to be stored. The dielectric material also helps to prevent the charges on the plates from escaping, thus maintaining the stored electrostatic energy.

## 5. Can the electrostatic energy in a capacitor be completely discharged?

Yes, the electrostatic energy in a capacitor can be completely discharged by allowing the charges on the plates to flow through a circuit. This can be done by connecting the two plates of the capacitor with a conducting material such as a wire. Once the charges have equalized, there will be no more electrostatic energy remaining in the capacitor.

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