# Finding potential between identical spheres

• flip290
In summary, the problem involves two oppositely charged identical insulating spheres with a diameter of 56.0cm and a uniform charge of magnitude 165μC, placed 1.20m apart center to center. The question asks for the potential difference between the nearest points on the surfaces of the spheres. To solve this, we need to find the individual potentials at points a and b near the surfaces of the spheres. Using the equation V=k(q/r), the potential at point a is found to be 1.002*106V, and the potential at point b would be the same but with the opposite sign. To find the potential difference, we need to take into account the contributions from both charges at each point.
flip290

## 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?

## 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.

Start by finding the individual potentials at points a and b near the surfaces of the spheres.

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?

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.

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?

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:
The difference would then be zero?

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?

## 1. What is the definition of potential between identical spheres?

The potential between identical spheres refers to the electric potential energy that exists between two spheres with equal charges. It is a measure of the work required to bring two identical charged spheres from infinity to a specific distance apart.

## 2. How is the potential between identical spheres calculated?

The potential between identical spheres can be calculated using the formula V = k(q/r), where V represents the potential energy, k is the Coulomb's constant, q is the charge of the spheres, and r is the distance between them.

## 3. What factors affect the potential between identical spheres?

The potential between identical spheres is affected by the magnitude of the charges on the spheres, the distance between them, and the medium in which they are placed. The potential will decrease as the distance between the spheres increases and increase as the charges on the spheres increase.

## 4. How does the potential between identical spheres relate to electric fields?

The potential between identical spheres is directly related to the electric fields that exist between them. The electric field is the force per unit charge acting on a charged particle, and the potential is the energy per unit charge at a specific point in the electric field.

## 5. What is the significance of finding the potential between identical spheres?

Finding the potential between identical spheres is important in understanding the behavior of electric charges and their interactions with each other. It also has practical applications in fields such as electrical engineering and physics, where knowledge of the potential between spheres can be used in designing and analyzing circuits and electric systems.

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