# A Question Related to Electrostatic Energy and Capacitors

## 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 Related Introductory Physics Homework Help News on Phys.org
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

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 ;)