Example: Getting forces from PE  3D
Prerequisites
 Energy of place  potential energy
 Forces from potential energy
 The gradient  a vector derivative
Understanding the situation
3D potential energies are of considerable importance in understanding electrical forces between atoms and molecules. But showing a 3D potential energy map is hard on a 2D surface like this computer screen. But you can get the idea from various 2D projections such as the one shown in the figures below. These represent a slice of a 3D situation with two lines of charge (one + ,one ). In this case, the "lines" of equal potential become surfaces. Imagine the loop rotated around a vertical axis out of the page to form distorted cylinders. The force associated with the change in the potential points perpendicular to these surfaces.
If we had an equation for the potential energy, $U(x,y,z)$, we could find the force it corresponds to using the vector derivative (gradient) equation:
$$\overrightarrow{F} = \overrightarrow{\nabla}U^{grav} =  \bigg(\frac{\partial U^{grav}}{\partial x} \hat{i} + \frac{\partial U^{grav}}{\partial y} \hat{j} + \frac{\partial U^{grav}}{\partial z} \hat{k}\bigg)$$
In this case we don't have the function so we have to do it qualitatively. The force is associate with the change in the potential and points in the direction of greatest change (downhill), perpendicular to the surfaces of equal potential. Where the surfaces are closest together (changing the fastest as a function of position), the force is the greatest.
Presenting a sample problem
Problem 1:
Where would a test charge feel the strongest electric force?

 A
 B
 C
 A and C
 It would not feel a force at any of the three points
Problem 2:
Where would a test charge feel the strongest electric force?

 A
 B
 C
 A and C
 It would not feel a force at any of the three points
Solving this problem
Problem 1: B
Don't get confused! The force is associated with the derivative of the PE  how fast it's changing, not what value it has. So even though the value of the PE is 0 at B, the lines are closest together there so the PE is changing the fastest. (Note that although the PE is 0 at B, it is positive just to the left and negative just to the right so it's definitely changing as the value passes through 0.)
Problem 2: B
Still B! That's where the lines are closest together so that's where the change is the greatest.
Last Modified: February 24, 2019