Charge of 2 conducting spheres separated by a distance

Click For Summary
The discussion centers on calculating the charge of two conducting spheres, one at a potential of 1500 V and the other at -1500 V, separated by 10.0 m. Initially, the charge on one sphere is calculated to be approximately 2.50×10−8 C, with a potential drop of 22.5 V at a distance of 10 m from the sphere. This drop is considered a small perturbation, allowing the assumption that the total potential can be approximated by adding this perturbation to the original 1500 V. The perturbation indicates that the influence of the second sphere is minimal, leading to a slight adjustment in the charge calculation to 2.54×10−8 C. The discussion highlights the relationship between potential, distance, and charge in electrostatics.
jolly_math
Messages
51
Reaction score
5
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.
 
Physics news on Phys.org
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%.
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?