Find field yielding the potential ln(1/r)

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SUMMARY

The discussion focuses on calculating the electric field strength from the given potential function, \(\Phi(\bar{r}) = \Phi_0\ln{\left(\frac{r_0}{r}\right)}\). The electric field is derived using the equation \(\bar{E}(\bar{r}) = -\nabla \Phi(\bar{r})\), leading to the expression \(\bar{E}(\bar{r}) = \Phi_0 \frac{1}{r} \hat{r}\). However, the expected answer is \(\Phi_0 \frac{1}{r^2} \hat{r}\), indicating a misunderstanding in the application of the gradient in spherical coordinates. The error stems from not correctly applying the gradient operator to the logarithmic potential.

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timn
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Dear Physics Forums,

Homework Statement



Given the potential

[tex]\Phi(\bar{r}) = \Phi_0\ln{\left(\frac{r_0}{r}\right)}[/tex]

calculate the electric field strength.

Homework Equations



[tex]- \nabla \Phi = \bar{E}[/tex]

The Attempt at a Solution



Due to the symmetry of the problem, consider the position vector [tex]\bar{r}[/tex] as a vector in spherical coordinates.

[tex]\bar{E}(\bar{r}) = -\nabla \Phi(\bar{r})[/tex]

The potential depends only on the radial component, so the other terms of the gradient vanish.

[tex]\bar{E}(\bar{r}) = - \Phi_0 \frac{\partial}{\partial r} \ln{\frac{r_0}{r}} \hat{r}<br /> = \Phi_0 \frac{1}{r} \hat{r}[/tex]

where [tex]\hat{r}[/tex] denotes the normalised radial base vector.

The answer is supposed to be [tex]\Phi_0 \frac{1}{r^2} \hat{r}[/tex], with a square in the denominator. What did I do wrong?
 
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"rhat" in the answer means the position vector, independent of the coordinate system.

ehild
 
Aha! Thank you, ehild.

Sorry about copying the answer wrong.
 

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