Electric potential constant within a conductor?

  • Thread starter Plamo
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1. Electric potential inside a conductor / outside a coaxial cable

Electric Potential inside a conductor(spherical) is a constant, although electric field is zero. How does that make sense given:
Given [itex]V=- \int E \cdot dl[/itex]?
The integral should be 0. Even if you consider constants of integration, shouldn't they drop off because the integral is from the radius to 0?

Given that potential is non-zero inside a conductor, does the same hold true outside a coaxial cable? A Gaussian surface around the cable shows that the electric field outside the cable is 0. Do we have the same case where the potential is non-zero outside of the cable?

Homework Equations


[itex]V=- \int E \cdot dl[/itex]


The Attempt at a Solution


The problem statement is my attempt at the solution. More of a lack of confusion than an actual problem.

Edit:
To clarify, this makes sense in reverse: E = del(V). Derivative of a constant is 0. How did that constant get there in the first place though?
 
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Answers and Replies

  • #2
Matterwave
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That integral is certainly 0 within the conductor. With the limits of integration:

[tex]V_2-V_1=\int__1^2 \vec{E}dl=0[/tex]

Obviously it's true since E is 0 inside the conductor...therefore the potential must be constant inside.

I guess I don't quite get what the problem is.
 
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The issue is that I don't see why it's non-zero inside. More importantly, I can't decide on whether or not it's 0 outside a coaxial cable.
 
  • #4
Matterwave
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What's non-0 inside? The electric field is definitely 0 inside a conductor (for electro-statics anyways).

How is your coaxial cable set up? Current moving in one direction inside and current moving in the opposite direction outside?
 
  • #5
jambaugh
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Consider two thick conducting plates connected to a battery so they have distinct constant potentials. If one is at potential zero the other is certainly not zero.

The potential equation you've given is more properly written:
[tex] \Delta V = V_2 - V_1 = -\int_{p_1}^{p_2} E\cdot dl[/tex]
In short it defines a potential difference.
 

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