Finding potential between identical spheres

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SUMMARY

The discussion focuses on calculating the electric potential between two oppositely charged identical insulating spheres, each with a diameter of 56.0 cm and a charge of 165 μC, positioned 1.20 m apart. The participants initially assumed that the potential difference would be zero due to equal and opposite charges, but this was incorrect. The correct approach involves calculating the individual potentials at points a and b on the surfaces of the spheres, considering contributions from both charges at each point. The final calculations indicate that the potential difference is not zero, and participants are encouraged to re-evaluate their calculations for accuracy.

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Homework Statement


Two oppositely charged identical insulating spheres, each 56.0cm in diameter and carrying a uniform charge of magnitude 165μC , are placed 1.20m apart center to center.
If a voltmeter is connected between the nearest points (a and b) on their surfaces, what will it read?
a1r0h1.jpg

Homework Equations


V=k((q1/r1)-q2/r2)

The Attempt at a Solution


At first I thought that the potential between the two would be zero because they are equal but opposite. However, that was incorrect. I have let r=0.6 and r=0.88 however, using both was still incorrect. I do not know where to go from here.
 
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Start by finding the individual potentials at points a and b near the surfaces of the spheres.
Show your work.
 
Va=k(q/r)=(8.99*109)((165*10-6)/.28)=5.298*106

The potential for Vb would be the same but opposite. Is this correct?
 
flip290 said:
Va=k(q/r)=(8.99*109)((165*10-6)/.28)=5.298*106

The potential for Vb would be the same but opposite. Is this correct?

What about the contribution of the field of the other sphere at a? Same goes for point b where the fields from both spheres contribute to the total.
 
This is where I am confused. Would you use the same equation but make r=1.20m?
 
flip290 said:
This is where I am confused. Would you use the same equation but make r=1.20m?

The distance is always from the center of the charge to the location in question.
 
So
Va=k(q/r)=(8.99*109)((165*10-6)/1.48)=1.002*106?
and the same for Vb?
 
flip290 said:
So
Va=k(q/r)=(8.99*109)((165*10-6)/1.48)=1.002*106?
and the same for Vb?

How did you arrive at the distance of 1.48m ? Also, there should be contributions from both charges.
 
The distance between the two is r=1.2m plus from the center of the sphere is 0.28m. Therefore the total distance would be 1.48m. If there are contributions from both charges am I not using the right equation?
 
  • #10
The distance between the spheres is specified center-to-center.

For a given location in space you need to find the contribution of both charges. That means applying your equation twice for each location, once for each charge that contributes.
 
  • #11
So would it be Va=(8.99*109)((165*10-6)/1.76) plus the contribution from Vb therefore the answer would be V=1.7*106V?
 
  • #12
flip290 said:
So would it be Va=(8.99*109)9((165*10-6)/0.88) plus the contribution from Vb therefore the answer would be V=3.37*106V?

I think you meant 0.28 rather than 0.88 in the above, as your result value looks okay.

That takes care of the potential at point a. As you said previously, the potential at b will be the same but with opposite sign. What then is the potential difference between the two points?

EDIT: I take it back, your result looks a bit low. Re-check your calculations.
 
Last edited:
  • #13
The difference would then be zero?
 
  • #14
flip290 said:
The difference would then be zero?

I just realized that your result for the potential at point a still looks a bit off (a bit low). Recheck your calculations. What are the individual contributions from the two charges at point a?

Regarding the difference value, if the potentials have opposite signs, how can the difference be zero?
 

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