Electric Potential at Radius R of Concentric Spheres

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Homework Help Overview

The discussion revolves around the electric potential at a specific radius of concentric spheres, particularly focusing on the relationship between charge and potential. Participants are exploring the implications of the formula V=kQ/R and the conditions under which it applies.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand why the charge is considered as 2Q in certain contexts and questioning the implications of being inside the outer shell. There are inquiries about the potential at various points and the effect of moving through the outer shell.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some have suggested that the potential remains constant when moving through the outer shell, while others are clarifying the limits of integration for calculating potential. There is a recognition of the need to clarify the total charge and distance involved in the calculations.

Contextual Notes

There appears to be some confusion regarding the definitions of distance and charge in the context of the problem. Participants are also navigating the implications of integrating the electric field to find potential, indicating a need for further exploration of these concepts.

lorx99
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Homework Statement



upload_2018-10-4_21-12-24.png

Homework Equations


V=kQ/R

The Attempt at a Solution


The answer is B)kQ/R. It is because V= k(2Q)/R. I don't understand why Q=2Q in this case. Isn't the point on the inside of the outer shell, so the Q for the equation is just Q?
 

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lorx99 said:

Homework Statement



View attachment 231679

Homework Equations


V=kQ/R

The Attempt at a Solution


The answer is B)kQ/R. It is because V= k(2Q)/R. I don't understand why Q=2Q in this case. Isn't the point on the inside of the outer shell, so the Q for the equation is just Q?
The potential at any point between infinity and the outer side of the outer shell is what?
And going thru the outer shell itself to the inner side of the outer shell changes that potential by how much?
 
rude man said:
The potential at any point between infinity and the outer side of the outer shell is what?
And going thru the outer shell itself to the inner side of the outer shell changes that potential by how much?
Potential from any point between inifitiy and outer side is just V=KQ/R. Going thru the outshell to just inner side of outershell is practically zero? So, is that why it is just V=KQ/R?
 
lorx99 said:
Potential from any point between inifitiy and outer side is just V=KQ/R. Going thru the outshell to just inner side of outershell is practically zero? So, is that why it is just V=KQ/R?
Edit: your answer is right but you reasoned wrong.
What is the distance in this case? What is the total charge?
 
Last edited:
rude man said:
Edit: your answer is right but you reasoned wrong.
What is the distance in this case? What is the total charge?
I am confused. What distance you mean?
 
lorx99 said:
I am confused. What distance you mean?
The distance d in the formula for potential kQ/d. Or, to find the potential by integrating the E field, what are the limits of integration?
 
rude man said:
The distance d in the formula for potential kQ/d. Or, to find the potential by integrating the E field, what are the limits of integration?
I think i understand. First, integrate from infinity to 2R which is K2Q/(2R). Then integrate from outer to inner surface of the outer shell, it would be 0 because E=0 inside the shell since V integeates over the efield which is just zer0. So the potential is KQ/r since the Qs cancel.
 
lorx99 said:
I think i understand. First, integrate from infinity to 2R which is K2Q/(2R). Then integrate from outer to inner surface of the outer shell, it would be 0 because E=0 inside the shell since V integeates over the efield which is just zer0. So the potential is KQ/r since the Qs cancel.
Straight A!
 

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