Charge of 2 conducting spheres separated by a distance

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SUMMARY

The discussion focuses on the electrostatic interaction between two identical conducting spheres, each with a radius of 15.0 cm, separated by a distance of 10.0 m. The charge on one sphere at a potential of +1500 V is calculated using the formula q = 4πεrV, resulting in a charge of 2.50×10−8C. The potential from the sphere at a distance of 10.0 m is determined to be 22.5 V, which is considered a small perturbation compared to 1500 V, allowing for the adjustment of the charge to 2.54×10−8C by treating the influence of the second sphere as a minor correction.

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  • Knowledge of the formula for charge calculation: q = 4πεrV
  • Basic grasp of perturbation theory in physics
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  • Study the derivation and applications of the formula for electric potential V = kq/r
  • Explore the concept of perturbation in electrostatics and its implications
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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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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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