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Some more trouble with Electric Charge Problems

  • Thread starter frankfjf
  • Start date
  • #1
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I've never been good at "Find the maximum whatever" (Area, charge in this case, etc...) type problems, so I could use some coaching on this one.

Homework Statement



Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q will the electrostatic force between the two spheres be maximized?

Homework Equations



q = Q/2, Fmax = kQ^2/4r^2

The Attempt at a Solution



Admittedly it's been pretty rough going on this one, and it's only the 3rd homework problem of the semester... but I've resolved to get through it all somehow. Through using the text's website companion I've gotten it down to the final step, which is where I get stumped. I've got it all down to:

k/r^2(qQ - q^2) = (3/4) (kQ^2/4r^2)

At this point the interactive help says all I gotta do is solve for q. But I can't seem to get q by itself. How can I manipulate the equation in order to isolate q? I tried a few things, such as dividing both sides by k to eliminate k, then multiplying both sides by r^2 to eliminate r^2. However, at that point I get stumped and have no idea how to further simplify the equation. Any help would be appreciated.
 
Last edited:

Answers and Replies

  • #2
486
1
From where you are:
[tex]\frac{1}{qQ-q^2} = \frac{3Q^2}{4}[/tex]

Clear all fractions (by multiplying both sides by each denominator), then get your equation in the form ax+bx+c=0 so you can use the quadratic equation.
 
  • #3
168
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Oh, I'm sorry, I wrote down my stumped point incorrectly. It's supposed to be this:

k(qQ - q^2)/r^2 = (3/4) (kQ^2/4r^2)

After performing the two steps I tried I'm left with:

(qQ - q^2) = 3Q^2/16
 
  • #4
486
1
OK. You can still multiply both sides by 16, then get your equation into the form ax^2+bx+c=0.

Oops. Sorry, last post I forgot to write squared after the first term.
 
  • #5
168
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I tried doing that but I'm still confused. Now I've got -16q^2 + 16qQ -3Q^2.

I'm not sure if that's the correct next step, because when I try to factor it I get something like (-4q - 3)(4q + 1). Plugging in I end up with -(3/4)Q and -(1/4)Q. Which is the answer except the answer isn't negative.
 
  • #6
486
1
I tried doing that but I'm still confused. Now I've got -16q^2 + 16qQ -3Q^2.
All of which =0.

I'm not sure if that's the correct next step, because when I try to factor it I get something like (-4q - 3)(4q + 1).
Actually, you get (-4q-3Q)(4q+Q)=0.

plugging in I end up with -(3/4)Q
I get positive 3/4

and -(1/4)Q. Which is the answer except the answer isn't negative.
Hmmm... which was the answer? 1/4 or 3/4?
 
  • #7
168
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Ah you are right on the positive (3/4)Q. It's listing both as the answer, but listing both as positive. The (1/4)Q still ends up negative on my end.
 
  • #8
486
1
I also get -1/4.

Conceptually, that makes no sense at all: if you transfer charge from one object to another, how can the charge change sign?

Mathematically, I just don't know how to get that -1/4 to be positive, or if you can just ignore the negative, or what.
 
  • #9
168
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Well, that I can clarify with my professor. Thanks for your help!
 

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