Electric potential

  • #1

Main Question or Discussion Point

In Griffith's electrodynamics chapter 2 example 7 he calculates potential due to a spherical shell outside it. Here E is radially outward while dr is radially inward as we are going towards the sphere hence E.dr should be negative but it is taken positive?
 

Answers and Replies

  • #2
sophiecentaur
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I don't have a copy of that book to hand but it is pretty likely that, if you are strict about the signs in your interpretation of the derivation, the 'correct' answer will emerge. Lines pointing away from the centre implies a positive charge on the sphere and Potential is defined as the Work done in bringing a Posirive charge from infinity to a point. You would be moving a positive charge towards a positive charge and that would involve Positive Work against the repulsive force. That's what we would expect. I think.
 
  • #3
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Here E is radially outward while dr is radially inward
Hmm, are you sure about that? I have never seen an example where the radial coordinate is inward pointing. The radial coordinate is universally 0 at the center and increasingly positive as you go further away from the origin. I have never once seen a radial coordinate which is 0 at the center and increasingly negative as you go further away.
 
  • #4
Hmm, are you sure about that? I have never seen an example where the radial coordinate is inward pointing. The radial coordinate is universally 0 at the center and increasingly positive as you go further away from the origin. I have never once seen a radial coordinate which is 0 at the center and increasingly negative as you go further away.
Doesn't dr points towards center as we move test charge towards the center? Thus dr is dr(-r^)
 
  • #5
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Doesn't dr points towards center as we move test charge towards the center? Thus dr is dr(-r^)
No, dr points in the direction of increasing r, by definition. If you are moving from large r to small r then that is reflected in your limits of integration.
 
  • #6
sophiecentaur
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Doesn't dr points towards center as we move test charge towards the center? Thus dr is dr(-r^)
The negative sign comes in the limits of the integration process. The direction of dr is taken care of by how the integration is performed. In the case of the potential around a sphere, it's fairly obvious but you need to stick to the integration rules for unusual field arrangements.
 
  • #7
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Doesn't dr points towards center as we move test charge towards the center? Thus dr is dr(-r^)
According to the definition of ## d\vec{r} ## used by Griffith (he actually uses the notation ## d\vec{l} ## for the displacement vector)
## d\vec{r} =dr\hat{r} ## .
This is valid for the radial displacement only. In general there are two more terms, for the angular displacement in spherical coordinates.
So the direction of the radial displacement is given by the sign of dr. In general it can have any direction.
The electric field is given by ## \vec{E}=\frac{q}{4 \pi \epsilon_0 r} \hat{r} ##
So the term ##\vec{E} \cdot d \vec{r} ## is
##\frac{q}{4 \pi \epsilon_0 r} dr ## as the product of the radial unit vectors is 1.
So you don't need to add a minus sign. The sign of the expression to be integrated is determined by the sign of dr. It can be positive or negative, depending on the direction of the integration. You don't need to worry about it, the sign of the result of the integral will be the one corresponding to the chosen direction of integration.
In the book example the integral is from infinity to some finite value. So you will integrate a negative quantity and the result will be negative. And with the minus in front you get positive value of the potential.
 
  • #8
sophiecentaur
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The sign of the expression to be integrated is determined by the sign of dr.
I guess that's a fairly basic thing because Potential is a scalar.
 

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