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QUESTION: Two point charges (+10.5 nC and -10.5 nC) are located 7.94 cm apart. Let U = 0 when all of the charges are separated by infinite distances. What is the potential energy if a third point charge q = -5.0 nC is placed at point b?

so im letting q1 = q2 = q = 10.5 nC, d = 3.97 cm and q3 = -5.0 nC
Im also using this equation: U = k(q1q2 / r12 + q1q3 / r13 + q2q3 / r23) and im still not getting the right answer. I'm not sure if im plugging in the wrong numbers. Please provide me with any help you can give. Thank you!

ANSWER: Your issue here is in the sign convention I think:  q1 is not equal to q2. Here, q1 = -q2.

So, without b being known and assuming it is someplace between the first two charges (which I will call points a=0 cm and c = 7.94 cm) the equation should look like this:

with a as a positive charge and b and c as negative we have something like


distances in short hand
r(ab) = b
r(ac) = c
r(bc) = c-b

U = k  ( qa * qb / b + qa * qc / c + qb * qc / (c-b) )

note that in the brackets the first term for ab interactions is negative overall (attractive), the second term is negative over all (attractive) and that the third term is positive overall (repulsive).

I hope this helps.

Best regards and good luck.

---------- FOLLOW-UP ----------

QUESTION: Hello Again,

Thanks so much for the reply. So I went ahead and tried modifying my answer and I set it up like this: U = (9*10^9) ((10.5 * 10^9) (-10.5 * 10^-9  /  (0.119 m) + (10.5 * 10^-9 * -5.0 * 10^-9) / (0.0794) + (-10.5 * 10^-9) ( -5.0 * 10^-9) / 0.0794)) but I know I'm doing something wrong. Could you please help me figure out what went wrong here?

Thank you so so much!

In your original question you stated that the third charge is placed at point b, but did not provide the actual distance or location of b.

I am working off of the assumption that b was placed directly between the first and second charge.  You will need to adjust your distances accordingly if this is not the case.
Also remember that r for these systems is the total distance between charges, not half.

r(ab) = 3.47 cm
r(ac) = 7.94 cm
r(bc) = 3.47 cm

U = k  ( qa * qb / r(ab) + qa * qc / r(ac) + qb * qc / r(bc) )
= 9E9 N*m^2/C^2 * ([ -10.5E-9 C * -5E-9 C / 0.0347 m ] + [ -10.5E-9 C * 10.5E-9 C / 0.0794 m ] + [ 10.5E-9 C * -5E-9 C / 0.0347 m ])


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Dr. Jeffery Raymond


Materials chemistry. Materials science. Spectroscopy. Polymer science. Physical Chemistry. General Physics. Technical writing. General Applied Mathematics. Nanomaterials. Optoelectronic Behavior. Science Policy.


Teaching: General Inorganic Chemistry I & II, Organic Chemistry I & II, Physical Chemistry I, Polymeric Materials, General Physics I, Calculus I & II
My prior experience includes the United States Army and three years as a development chemist in industry. Currently I am the Assistant Director of the Laboratory for Synthetic Biological Interactions. All told, 13 years of experience in research, development and science education.

Texas A&M University, American Chemical Society, POLY-ACS, SPIE

Journal of the American Chemical Society, Nanoletters, Journal of Physical Chemistry C, Journal of Physical Chemistry Letters, Ultramicroscopy Proceedings of SPIE, Proceedings of MRS, Polymer News, Chemical and Engineering News, Nano Letters, Small,, Angewandte

PhD Macromolecular Science and Engineering (Photophysics/Nanomaterials Concentration), MS Materials Science, BS Chemistry and Physics, Graduate Certificate in Science Policy, AAS Chemical Technology, AAS Engineering Technology

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