A Question Related to Electrostatic Energy and Capacitors

  • Thread starter Air_Gear
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  • #1
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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 :wink:) 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 :smile:
 

Answers and Replies

  • #2
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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.
 
  • #3
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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
 
  • #4
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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 ;)
 

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