Electrostatics - metal sphere potential

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Homework Help Overview

The discussion revolves around the electrostatics of two metal spheres, one initially charged to 10V and then connected to another sphere of double the radius. Participants are exploring the implications of charge distribution and potential changes upon connection.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate electric field and potential to derive the final potential when the spheres are connected. Some participants question the validity of the initial approach and suggest using charge distribution principles instead.

Discussion Status

Participants are actively engaging with the problem, with some providing alternative methods to calculate the final potential. There is a mix of interpretations regarding the relationship between charge and potential, and while some calculations are presented, there is no explicit consensus on the correct approach yet.

Contextual Notes

There is an underlying assumption about the conservation of charge and the requirement for equal potential across connected conductors. The original poster's calculations are based on their understanding of electric fields and potentials, which some participants challenge.

xwhyy
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Hi,

again very simple problem

Homework Statement



Metal sphere of radius R1 is charged to 10V.
It is then connected to another sphere with R2 = 2*R1.
What is the final potential?

Homework Equations





The Attempt at a Solution



V = E * distance
so in the case of disconnected it is E*R1
when connected R's sum up so it would be E*(R1+R2)
and since I don't know E but R2=2*R1
it is E*3*R1 so my solution is:

V_connected = V * 3

Is that correct?
 
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If it's correct, that's only a coincidence.

What is the potential of a sphere of radius R1 and charge Q?

Set that equal to 10V and solve for Q.

Then distribute the charge over the two spheres so that their potentials are equal.
 
Thank you for your reply!

so:

V = Q / [4*pi*epsilon*r1]
Q = V * 4* pi * epsilon * r1

V_connected = (V * 4 * pi * epsilon * r1) / ( 4 *pi * epsilon * 2 * r1)
which after simplification is equal
V_connected = V/2

is that true? :)
 
hop hop
 
xwhyy said:
...

V = Q / [4*pi*epsilon*r1]
Q = V * 4* pi * epsilon * r1
Looks good. BTW: V = 10 Volts

When they're connected: V1 = V2 and Q1 + Q2 = Q = V·4πε0·R1.
 

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