- #1

timn

- 19

- 0

## 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}

= \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?