Gradient of a potential energy function

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
Radarithm
Gold Member
Messages
158
Reaction score
2

Homework Statement


Find the derivative of [tex]\frac{Q}{4\pi \epsilon_0 r}[/tex]


Homework Equations


[tex]\frac{d}{dx} \frac{1}{x}=\ln x[/tex]


The Attempt at a Solution



Assuming [itex]Q[/itex] and the rest of the variables under it are constant, [tex]\frac{Q}{4\pi \epsilon_0}\frac{1}{r}[/tex] then the derivative should be [itex]\ln r[/itex]. I am taking the gradient of a potential energy function but since it is in one dimension ([itex]r[/itex] 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?
 
Physics news on Phys.org
Mentallic said:
You have it the wrong way around.

[tex]\frac{d}{dx}\ln{x}=\frac{1}{x}[/tex]

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