Two-electrode spherical system (Potential)

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  • #1
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Homework Statement



See figure attached.

Homework Equations





The Attempt at a Solution



See figure attached for the provided solution.

I got everything up to what I put in a red box.

Where does he get that negative from?

Did he do that with the intentions of reversing the limits on his integral?
 

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  • #2
SammyS
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[itex]\displaystyle \Phi_r-\Phi_b=-\int_b^r\vec{E}\cdot\vec{dr}[/itex]
 
  • #3
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[itex]\displaystyle \Phi_r-\Phi_b=-\int_b^r\vec{E}\cdot\vec{dr}[/itex]
Isn't that the potential at some radius r, with the reference set at radius b?

In the question it says that the reference for the potential should be the potential on the inner conductor where the radius is a.

I have the urge to write,

[tex]\phi(r) = \int_{b}^{r} \vec{E} \cdot \vec{dr} + C[/tex]

Where C is chosen such that,

[tex]\phi(a) = 0[/tex]

but we already know from the first part that it is already 0, so C=0.
 
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  • #4
SammyS
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From the first part, we know that Φb = Φa
 
  • #5
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From the first part, we know that Φb = Φa
On the third line from the bottom, didn't he miss a negative when he evaluated his integral?

It should be,

[tex]\left(-\frac{1}{r} \right)^{r}_{c} = \frac{1}{c} - \frac{1}{r}[/tex]

Correct?
 
  • #6
SammyS
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After looking at it for some time, I agree with you.
 

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