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Charge on a sphere.

  • Thread starter SheldonG
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Homework Statement


Two identical small conducting spheres are separated by 0.60 m. The spheres carry different amounts of charges and each sphere experiences an attractive electric force of 10.8 N. The total charge on the two spheres is [itex]=-12\mu C[/itex]. What is the positive charge on one of the spheres?


Homework Equations


Coulomb's Law.


The Attempt at a Solution


Part of my problem is I think I am confused about how to treat the signs of the charges.

[tex] q_1 + q_2 = -12 \times 10^{-6} [/tex]
[tex] q_1 = -(q_2 + 12) [/tex]

Coulomb's Law gives:

[tex] 10.8 = k\frac{q_2(q2_+12)}{0.60^2} [/tex]

I left out the negative sign, because of the absolute value in Coulomb's law.
Is this correct?

and leads me to

[tex] 0 = q_2^2 + (12\times 10^{-6})q_2 - 4.32601 \times 10^{-10} [/tex]

From this, I end up with two solutions that fit the conditions of the problem. Is that right? I get a positive root of 27.6 micro C, and a negative root of 15.6 micro C, which gives me the positive charge of 3.6 micro C.

Could anyone confirm that this is the right way to handle this situation, or clear up my sign confusion for me?

Thanks so much,
Sheldon
 
Last edited:

Answers and Replies

  • #2
Hootenanny
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Indeed, the signs seems to be your problem, the first line should read

[tex] q_1 + q_2 = 12 \times 10^{-6} [/tex]
 
  • #3
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Hi Hootenanny,

Sorry, I made a typo when I entered the problem. The sum of the charges is -12 micro C. I'll edit the original post to fix that.

Sheldon
 
  • #4
Hootenanny
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Then your method looks good to me...:approve:
 
  • #5
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Thanks! So there are actually two possible pairs of charges for this situation? Interesting...
 
  • #6
nrqed
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Homework Statement


Two identical small conducting spheres are separated by 0.60 m. The spheres carry different amounts of charges and each sphere experiences an attractive electric force of 10.8 N. The total charge on the two spheres is [itex]=-12\mu C[/itex]. What is the positive charge on one of the spheres?


Homework Equations


Coulomb's Law.


The Attempt at a Solution


Part of my problem is I think I am confused about how to treat the signs of the charges.

[tex] q_1 + q_2 = -12 \times 10^{-6} [/tex]
[tex] q_1 = -(q_2 + 12) [/tex]

Coulomb's Law gives:

[tex] 10.8 = k\frac{q_2(q2_+12)}{0.60^2} [/tex]

I left out the negative sign, because of the absolute value in Coulomb's law.
Is this correct?

and leads me to

[tex] 0 = q_2^2 + (12\times 10^{-6})q_2 - 4.32601 \times 10^{-10} [/tex]

From this, I end up with two solutions that fit the conditions of the problem. Is that right? I get a positive root of 27.6 micro C, and a negative root of 15.6 micro C, which gives me the positive charge of 3.6 micro C.

Could anyone confirm that this is the right way to handle this situation, or clear up my sign confusion for me?

Thanks so much,
Sheldon
whatch out.. there must be something wrong (I will check in a second) because it's clear that with your two solutions, the product of the charges is not the same so both can't be solutions (I mean that 27.6 times (-39.6) is obviously not the same as -15.6 times +3.6) !!!
 
  • #7
nrqed
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Homework Statement


Two identical small conducting spheres are separated by 0.60 m. The spheres carry different amounts of charges and each sphere experiences an attractive electric force of 10.8 N. The total charge on the two spheres is [itex]=-12\mu C[/itex]. What is the positive charge on one of the spheres?


Homework Equations


Coulomb's Law.


The Attempt at a Solution


Part of my problem is I think I am confused about how to treat the signs of the charges.

[tex] q_1 + q_2 = -12 \times 10^{-6} [/tex]
[tex] q_1 = -(q_2 + 12) [/tex]

Coulomb's Law gives:

[tex] 10.8 = k\frac{q_2(q2_+12)}{0.60^2} [/tex]

I left out the negative sign, because of the absolute value in Coulomb's law.
Is this correct?
[/itex]
You have to be careful.

What they give is the magnitude, so you should really write
[tex] 10.8 = k \frac{|q_2| \times |q_2 +12|}{0.6^2} [/tex]

Now, what to do with the absolute values? the rule is

[tex] |q_2| = q_2 ~if~ q_2 >0 [/tex]
[tex] |q_2| = -q_2 ~if~ q_2 <0 [/tex]
[tex] |q_2+12| = q_2 +12~if~ q_2+12 >0 \rightarrow q_2> -12 [/tex]
[tex] |q_2| =- q_2 -12~if~ q_2 +12<0 \rightarrow q_2 < -12[/tex]


So there are several cases to consider. If q_2 > 0, then you may use [itex] |q_2| |q_2 +12| = q_2 (q_2+12)[/itex] like you did. this means that you must reject any solution with q_2 negative from your solution.
The solution of that quadratic are [itex] -27.6 \muC [/itex] or [itex]+15.6 \muC
[/itex] Then we keep only the +15.6 solution which means the other charge is -27.6 microcoulombs (this gives an attractive force as the question asks and this gives the correc magnitude of the force as can be checked.



Then you may consider -12< q_2 < 0 in which case [itex] |q_2| |q_2 +12| = - q_2 (q_2+12)[/itex] This gives you a different quadratic solution that you will solve and accept the solution only if q_2 is in the correct range. It turns out there is no solution.


Finally, if q_2 < -12, you have to use [itex] |q_2| |q_2 +12| = -q_2\times (-q_2-12)[/itex] which will give you back the same equation you had so the solutions will be the same but now you reject any solution that is not below -12. This leaves the solution of -27.6 microC so the other charge is +15.6 microC.

This makes sense...there are two solutions in the sense that either charge may be -27.6 microC with the other being +15.6 microC.


Patrick
 
  • #8
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Patrick,

Thank you. That was incredibly enlightening. I really appreciate it!

Sheldon
 

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