# Gradient of a potential energy function

1. Jan 26, 2014

1. The problem statement, all variables and given/known data
Find the derivative of $$\frac{Q}{4\pi \epsilon_0 r}$$

2. Relevant equations
$$\frac{d}{dx} \frac{1}{x}=\ln x$$

3. The attempt at a solution

Assuming $Q$ and the rest of the variables under it are constant, $$\frac{Q}{4\pi \epsilon_0}\frac{1}{r}$$ then the derivative should be $\ln r$. I am taking the gradient of a potential energy function but since it is in one dimension ($r$ in this case isn't a 2-3 dimensional vector) it is the same as taking the derivative. Is my answer correct or did I make a mistake somewhere?

2. Jan 26, 2014

### Mentallic

You have it the wrong way around.

$$\frac{d}{dx}\ln{x}=\frac{1}{x}$$

3. Jan 26, 2014

Yep, thanks for correcting me. The derivative is then $$\frac{-1}{r^2}$$ from the power rule, correct?