Electric Potential and Electric Field

Remembering that the electric field points downhill, consider the following problem.
The figure shows the electric potential along a line. Four points are marked on the line.
At which point or points does the electric field point to the left (in the \(-\hat x\) direction)?
The electric field points downhill. At point D, downhill is towards the left. (If you were tempted to say C as well, because the potential is negative there, remember that a different choice of \(V_{\infty}\) could make that potential zero or positive.)
At which point or points is the electric field zero?
Since point C is a minimum of the potential, there is no "downhill" at that spot, and so the electric field doesn't point to the left or right: the field is zero there.
Where is the electric field strongest?
This is a little handwavy, but if the electric field points downhill, it makes sense that it should be stronger at point A, the steepest point (the most "downhilliest point") of the graph.

From these examples, you've probably surmised that the electric field is related to the slope of the graph, or the derivative of the potential. But notice: the slope of the graph at point D is positive, but the electric field there points in the negative direction. Similarly, the slope of the graph at points A and B is negative, but $E_x>0$. So the electric field is actually related to the negative slope of the potential graph, and in fact it's equal to it.
$$E_x=-\frac{\partial V}{\partial x}$$
(If you've taken multivariable calculus you'll recognize the \(\partial\) symbol; if not, you can think of it as just a fancy way of writing \(d\) in a derivative. The distinction isn't terribly important for us.)
And since the electric field is the negative derivative of the potential, the potential must be the negative integral of the electric field.
$$V=-\int E_x\,dx$$
Remember how indefinite integrals have arbitrary constants? This is why electric potential has an arbitrary \(V_{\infty}\): it's actually an integral! The potential difference between two points \(x_A\) and \(x_B\) is $$\Delta V=V(x_B)-V(x_A)=-\int_{x_A}^{x_B}E_x\,dx$$
Consider this infinite plane, with a surface charge density \(\sigma\). Recall that it creates an electric field of \(2\pi k\sigma \hat x\).
If \(x=0\) at the surface, find the potential of the plane as a function of \(x\).
The \(x\) component of the electric field is \(E_x=2\pi k\sigma\). The electric potential is thus $$\begin{eqnarray} V&=&-\int E_x\,dx\\ &=&-\int 2\pi k\sigma\,dx\\ &=&-2\pi k\sigma x+C\\ \end{eqnarray}$$

What does \(C\) represent? Well, the potential is equal to \(C\) when \(x=0\), and so \(C\) is the electric potential at the infinite plane. Because the potential is linearly proportional to position, the equipotential surfaces are evenly spaced.

What happens if we set \(V_\infty=0\)?
If \(V=0\) at \(x=\infty\), the equation is $$0=-2\pi k\sigma\infty + C \implies C=\infty$$ If \(C\) is infinite, that means that the potential anywhere near the sheet is going to be infinite. This would be very impractical, so this is one case where definining \(V_\infty=0\) would be a bad idea.

Find the potential difference between these two points which are at an angle with respect to each other.