Charge of 2 conducting spheres separated by a distance

In summary: The 22.5V is related to the 1500V because the potential gradient is 1500V/0.150m.In summary, the charge on each sphere with a potential of +1500V and -1500V, separated by a distance of 10.0m, can be calculated using the equation q = 4πεrV. The potential from one sphere at a distance of 10.0m is 22.5V. This difference is considered a small perturbation and can be treated as a slight increase in both charges by 1.5%. Therefore, the charges on each sphere can be calculated as q = 4π(8.85×10−12C2
  • #1
jolly_math
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Homework Statement
Two identical conducting spheres of radius 15.0 cm are separated by a distance of 10.0 m. What is the charge on each sphere if the potential of one is +1500 V and the other is -1500 V? Take V = 0 at infinity.
Relevant Equations
V = E∆s
V = kq/r
First assuming only one sphere at a potential of 1500 V, the charge would be q = 4πεrV = 4π(8.85×10
−12C2/N · m)(0.150 m)(1500 V) = 2.50×10−8C.
The potential from the sphere at a distance of 10.0 m would be V =(1500V)(0.150m)/(10.0m) =22.5V.

I don't understand the reasoning of the following:

This is small compared to 1500V, so we will treat it as a perturbation. This means that we can assume that the spheres have charges of
q = 4πεrV = 4π(8.85×10−12C2/N · m)(0.150 m)(1500 V + 22.5 V) = 2.54×10−8C.

What does the perturbation refer to, and how is 1500 V + 22.5 V related to the specific distance 0.150 m? Thank you.
 
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  • #2
jolly_math said:
Homework Statement:: Two identical conducting spheres of radius 15.0 cm are separated by a distance of 10.0 m. What is the charge on each sphere if the potential of one is +1500 V and the other is -1500 V? Take V = 0 at infinity.
Relevant Equations:: V = E∆s
V = kq/r

First assuming only one sphere at a potential of 1500 V, the charge would be q = 4πεrV = 4π(8.85×10
−12C2/N · m)(0.150 m)(1500 V) = 2.50×10−8C.
The potential from the sphere at a distance of 10.0 m would be V =(1500V)(0.150m)/(10.0m) =22.5V.

I don't understand the reasoning of the following:

This is small compared to 1500V, so we will treat it as a perturbation. This means that we can assume that the spheres have charges of
q = 4πεrV = 4π(8.85×10−12C2/N · m)(0.150 m)(1500 V + 22.5 V) = 2.54×10−8C.

What does the perturbation refer to, and how is 1500 V + 22.5 V related to the specific distance 0.150 m? Thank you.
It just means that if the influence of the far sphere only drops the potential by 22.5/1500=1.5% then we can restore it to 1500V, near enough, by increasing both charges by 1.5%.
 
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1. How does the distance between two conducting spheres affect the charge on each sphere?

The charge on each sphere is directly proportional to the distance between them. As the distance increases, the charge on each sphere decreases. This relationship is known as Coulomb's law.

2. Can the charge on one sphere affect the charge on the other sphere?

Yes, the charge on one sphere can affect the charge on the other sphere due to the principle of electrostatic induction. If one sphere has a larger charge, it can induce a charge on the other sphere without direct contact.

3. What happens to the electric field between two conducting spheres when they are brought closer together?

As the distance between the spheres decreases, the electric field between them increases. This is because the closer the spheres are, the stronger the electric force between them.

4. How does the charge on each sphere affect the strength of the electric field between them?

The charge on each sphere has a direct impact on the strength of the electric field between them. The higher the charge on each sphere, the stronger the electric field will be between them.

5. Can the charge on one sphere be transferred to the other sphere?

Yes, if the two spheres are in contact, the charge can be transferred from one sphere to the other until they both have equal and opposite charges. This is known as charging by conduction.

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