- #1
timn
- 19
- 0
Dear Physics Forums,
Given the potential
\Phi(\bar{r}) = \Phi_0\ln{\left(\frac{r_0}{r}\right)}
calculate the electric field strength.
- \nabla \Phi = \bar{E}
Due to the symmetry of the problem, consider the position vector \bar{r} as a vector in spherical coordinates.
\bar{E}(\bar{r}) = -\nabla \Phi(\bar{r})
The potential depends only on the radial component, so the other terms of the gradient vanish.
\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}
where \hat{r} denotes the normalised radial base vector.
The answer is supposed to be \Phi_0 \frac{1}{r^2} \hat{r}, with a square in the denominator. What did I do wrong?
Homework Statement
Given the potential
\Phi(\bar{r}) = \Phi_0\ln{\left(\frac{r_0}{r}\right)}
calculate the electric field strength.
Homework Equations
- \nabla \Phi = \bar{E}
The Attempt at a Solution
Due to the symmetry of the problem, consider the position vector \bar{r} as a vector in spherical coordinates.
\bar{E}(\bar{r}) = -\nabla \Phi(\bar{r})
The potential depends only on the radial component, so the other terms of the gradient vanish.
\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}
where \hat{r} denotes the normalised radial base vector.
The answer is supposed to be \Phi_0 \frac{1}{r^2} \hat{r}, with a square in the denominator. What did I do wrong?
