Review / Chapter 2 — Vector Calculus

Equipotentials

Chapter 2 — Vector Calculus

An equipotential is a set of points where the electric potential \(V\) takes a single constant value. In 2D they are curves; in 3D they are surfaces. Sketching equipotentials together with field lines gives a complete picture of an electrostatic field — you can read off the direction of \(\vec{E}\) and estimate its magnitude just by looking.

Always perpendicular to field lines

Because \(\vec{E} = -\vec{\nabla} V\) and the gradient is perpendicular to level sets of \(V\), the electric field is always perpendicular to equipotentials. If \(\vec{E}\) had any component along an equipotential, moving along that direction would change \(V\), contradicting constancy. Equivalently: moving along an equipotential does no work on a test charge.

Spacing encodes field strength

If you draw equipotentials at evenly-spaced values of \(V\) (say every \(\Delta V = 1\text{ V}\)), the spacing between neighboring equipotentials tells you how strong \(\vec{E}\) is:

\[|\vec{E}| \approx \frac{\Delta V}{\Delta s},\]

where \(\Delta s\) is the perpendicular distance between adjacent equipotentials. Tightly packed equipotentials mean a strong field; widely spaced equipotentials mean a weak field.

Common examples

Reading a diagram

Given a drawing with equipotentials labeled \(V_1 > V_2 > V_3\):

  1. The field points from higher to lower \(V\), perpendicular to the curves, crossing them orthogonally.
  2. Where equipotentials crowd together, \(|\vec{E}|\) is largest.
  3. At a saddle between equipotentials, the field points along the direction of steepest potential decrease.

Note: a diagram that just shows unlabeled equipotentials is ambiguous about whether \(\vec{E}\) points radially in or radially out — you need to know which way \(V\) is increasing.

Practice Problems

Problem 1easy
A potential \(V(x,y,z) = -E_0 z\) describes a uniform field. Describe the equipotentials and the field.
Hint
Set \(V\) equal to a constant and solve for a surface.
Solution

\(V = \text{const}\) means \(z = \text{const}\), so equipotentials are horizontal planes. \(\vec{E} = -\vec{\nabla} V = (0,0,E_0)\), perpendicular to the planes.

Answer: horizontal planes; \(\vec{E} = E_0\hat{z}\).

Problem 2easy
Why can two equipotentials at different values of \(V\) never cross?
Hint
What would \(V\) be at the crossing point?
Solution

At a crossing, \(V\) would have two different values at the same point — impossible, since \(V\) is a single-valued scalar function of position.

Answer: different equipotentials can't cross because \(V\) at the intersection point would be ambiguous.

Problem 3medium
In a 2D diagram equipotentials are drawn at \(V = 5,4,3,2,1\) V, with the \(5\) V curve innermost. At point \(P\) adjacent equipotentials are \(0.2\) mm apart, and at point \(Q\) they are \(1.0\) mm apart. Compare \(|\vec{E}|\) at \(P\) vs \(Q\).
Hint
\(|\vec{E}| \approx \Delta V / \Delta s\) with \(\Delta V = 1\) V.
Solution

At \(P\): \(|\vec{E}_P| \approx 1/(0.2\times 10^{-3}) = 5000\) V/m. At \(Q\): \(|\vec{E}_Q| \approx 1/(1.0\times 10^{-3}) = 1000\) V/m.

Answer: \(|\vec{E}_P|\) is five times \(|\vec{E}_Q|\).

Problem 4medium
The potential \(V(x,y) = x^2 + y^2\) has equipotentials in the \(xy\)-plane. Describe them and find the field.
Hint
Level sets of \(x^2+y^2\).
Solution

\(x^2+y^2 = c\) for \(c \ge 0\) are circles of radius \(\sqrt{c}\). \(\vec{E} = -\vec{\nabla} V = (-2x,-2y,0) = -2\vec{r}\) in the plane — radially inward, with magnitude growing linearly in \(r\).

Answer: concentric circles; \(\vec{E} = -2(x,y,0)\).

Problem 5medium
Sketch (or describe) the equipotentials for two equal, opposite point charges \(\pm q\) separated by a distance \(d\). What is special about the symmetry plane midway between them?
Hint
By symmetry, the potential contributions cancel on the perpendicular bisector.
Solution

Equipotentials are lobed surfaces that close around each charge separately at small distances and distort into a dipole-like shape far away. The perpendicular bisector plane is the \(V=0\) equipotential: every point on it is equidistant from both charges, so the \(+q\) and \(-q\) contributions to \(V\) cancel exactly.

Answer: two sets of closed lobes around each charge, plus the \(V=0\) perpendicular bisector plane.