# Connection field lines/potential/vector field

#### Funzies

Hey guys,
I'm following a course on vector calculus and I'm having some trouble connecting things. Suppose we have a three-dimensionale vectorfield F(x,y,z)=(F1,F2,F3) and suppose we have a potential phi for this. So: F=grad(phi).
The field lines of a vector field are defined as d(r)/dt = lamba(t)F(r(t)). In words: the derivative of the field line (which is parametrized bij r) is parallel to F(r(t)).

Now for my question. Suppose we have equipotential lines, so phi=c with c a constant. What's the connection between these equipotential lines, the field lines and the vector field in terms of being parallel or right-angled?

I think the answer should be that the field lines are right-angled with the equipotential lines, but I don't know how to deduce this.

#### HallsofIvy

Homework Helper
Let f(x,y,z) be the value of the potential field at each point (x,y,z). Then the vector $grad f= \nabla f$ points in the direction of the "field lines", the lines of fastest increase of the function f. Further, the rate of increase of f in the direction of unit vector $\vec{v}$ is given by $\nabla f\cdot \vec{v}$. That is, the direction in which the derivative is 0, the "equipotential lines" (strictly speaking, in three dimensions, they would be equipotential surfaces) is exactly the direction in which that dot product is 0, the direction in which the vector $\vec{v}$ is perpendicular to $\nabla f$ and so perpendicular to the field lines.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving