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I was wondering if for a point-like charged object, does the gradient of the electric potential point in the direction of maximum increase or maximum decrease of the function $V$?

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The gradient points in the direction of maximum increase. The electric field, which is the negative of the gradient (and probably what you are asking about), points in the direction of maximum decrease. – DJBunk Mar 3 '13 at 15:24
If the sign of the charge changes, does this affect the direction ? – Carpediem Mar 3 '13 at 15:25
The electric field will always point in the direction of maximum decrease. This direction will depend on the sign of the charge, so they will be opposite for oppositely charged objects. But it still points in the direction of maximum decrease. – DJBunk Mar 3 '13 at 15:27
Can I ask you a final question. Does the electric field depend on my choice of point for the zero reference electric potential ? – Carpediem Mar 3 '13 at 15:28
You can ask all the questions you want. The electric field doesn't depend on your choice for zero potential since the electric field is the gradient of the potential. Only differences in potential energy are meaningful, and electric potential is just electric potential per unit charge, so only differences in electric potential are meaningful. – DJBunk Mar 3 '13 at 15:33
up vote 4 down vote accepted

I'd like to add a bit of mathematical detail the (correct) statements by DJBunk. Let a scalar function $f$ be given (let's not restrict ourselves to the electric potential). For any unit vector $\mathbf n$, we can define the directional derivative $D_\mathbf{n}$ of the function $f$ in the direction $\mathbf n$ as follows: $$ D_\mathbf{n}f(\mathbf x) = \mathbf n\cdot\nabla f(x). $$ The directional derivative gives the rate of change of the scalar function $f$ in the direction of the unit vector $\mathbf n$. Notice that $$ \mathbf n\cdot \nabla f(\mathbf x) = |\nabla f(\mathbf x)|\cos\theta $$ where $\theta$ is the angle between $\mathbf n$ and $\nabla f(\mathbf x)$, so the directional derivative is maximized when $\theta = 0$, and is minimized when $\theta = -\pi$. In other words;

The the rate of change of a scalar function $f$ at a point $\mathbf x$ is positive and greatest in magnitude in the direction of the gradient of $f$ at $\mathbf x$.

This confirms BJBunk's statements.

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+1 - Thanks for fleshing this out. – DJBunk Mar 4 '13 at 0:16
@DJBunk Ain't no thang. – joshphysics Mar 4 '13 at 1:50

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