# Charge of 2 conducting spheres separated by a distance

• jolly_math

#### jolly_math

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.

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%.

• jolly_math