Can Two Identical Charged Conductors Have a Potential Difference?

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SUMMARY

The discussion centers on the potential difference between two identical charged conductors. It is established that while two nearby conductors carrying the same negative charge can theoretically have a potential difference, this is contingent upon their geometrical shapes. If both conductors are identical in shape and charge, no potential difference exists due to the absence of an electric field between them. The capacitance formula, C = Q/V, becomes ineffective in this scenario as both Q and V approach zero.

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  • Understanding of electric fields and potential difference
  • Knowledge of capacitance and its formula, C = Q/V
  • Familiarity with conductor geometry and its impact on electric fields
  • Basic principles of electrostatics
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jezza10181
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Hi,

I found the following question in a physics book, and so dusted off my 30yr old knowledge on capacitors and tried to answer it. The question is as follows :-

"Suppose two nearby conductors carry the same negative charge. Can there be a potential difference between them? If so, can the definition of capacitance, C = Q/V, be used here?"

My own answer is that, yes, there can be a potential difference between the conductors, but a no to the second part. There surely cannot be a capacitance type effect between the conductors as the charges are the same, and so would repel each other and therefore not want to 'accumulate'.

What do you think?
 
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It would seem to me that each of the conductors has a capacitance, so that the voltage they each experience is proportional to the charge, (with a proportionality constant that is the inverse of the capacitance). The capacitance of the two conductors can be different, and thereby different voltages, even if the charge is the same.
 
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Likes Delta2, alan123hk, sophiecentaur and 1 other person
jezza10181 said:
"Suppose two nearby conductors carry the same negative charge. Can there be a potential difference between them? If so, can the definition of capacitance, C = Q/V, be used here?"

I think this https://www.comsol.com/blogs/how-to-calculate-a-capacitance-matrix-in-comsol-multiphysics/ is a general method to solve the capacitance problem of multi-conductor systems, but it is more complicated and abstract, and it is usually not used when dealing with simple daily engineering problems. If you are interested, you can refer to it.
 
jezza10181 said:
"Suppose two nearby conductors carry the same negative charge. Can there be a potential difference between them? If so, can the definition of capacitance, C = Q/V, be used here?"

My own answer is that, yes, there can be a potential difference between the conductors, but a no to the second part. There surely cannot be a capacitance type effect between the conductors as the charges are the same, and so would repel each other and therefore not want to 'accumulate'.

What do you think?
There is no potential difference between the two equally charged conductors since there is no E field between them (I am assuming two similar large plates.)

The formula C = Q/V is not wrong, it's just useless since both the numerator & denominator ## \rightarrow 0 ##.
 
It depends on the geometry (their shape basically) of the conductors, if both have identical geometry and the same charge then there will be no potential difference. However if the two conductors have different shapes, then there will be potential difference even if they have the same charge.
 
Delta2 said:
It depends on the geometry (their shape basically) of the conductors, if both have identical geometry and the same charge then there will be no potential difference. However if the two conductors have different shapes, then there will be potential difference even if they have the same charge.
Right, but I assumed two large (very large!) plates so fringing fields are negligible and the E field is zero "everywhere".
 

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