Charged Rod
z = ∫₋₁¹ 1/hypot(x − u, y) du
Open in the app The dials and keys named below are the app's.
What it draws
The height is a sum over a line of charge: z = ∫₋₁^1 du / hypot(x − u, y). A point charge at (u, 0) has the Coulomb potential 1/distance, in units that drop the constant, and hypot(x − u, y) is its distance from the point (x, y) of the sheet. Integrating u from −1 to 1 adds up a rod of length 2 along the x axis, one unit of charge per unit length. So the sheet is the electrostatic potential of a uniformly charged segment. There is no t: the field is static.
The integral has a closed form, asinh((1 − x)/|y|) + asinh((1 + x)/|y|). Above the middle of the rod at y = 1 it is 2 · asinh(1) ≈ 1.763. Far away the rod looks like a point charge of 2 and z ≈ 2/r: at (0, 3) the sheet reads 0.655 against 2/3 ≈ 0.667. Close in, z ≈ 2 · ln(2/|y|) grows without bound, 5.99 at y = 0.1. On the rod itself the integral is infinite. The grid line along y = 0 only meets that at the origin, where a sample lands on the pole and the sheet has a hole. Elsewhere on the rod it reads 11 to 23, which the height clamp draws as a flat-topped wall.
The shape
The level curves of the sheet are ellipses with their foci at the ends of the rod, (−1, 0) and (1, 0). On the ellipse with semi-major axis a the potential is ln((a + 1)/(a − 1)) , ln 3 ≈ 1.099 on the one through (2, 0). Near the rod the contours hug it; far out they round into circles. The rod is the degenerate ellipse in the middle, the ridge line. Its ends stand lower than its middle, 3.69 at (1, 0.1) against 5.99 above the centre, since charge lies on one side only.
History
Charles-Augustin de Coulomb measured the inverse-square force between charges with a torsion balance in 1785. The potential, one number at each point whose slope is the field, took its name from George Green's essay of 1828.
Try
- Top: the contours are the confocal ellipses, thin round the rod and nearly circular far out.
- Lengthen the rod, integral(1/hypot(x − u, y), u, −2, 2): twice the charge, so the far field is 4/r and the foci move to (±2, 0).
- Set Mesh to an odd count such as 45: no grid line runs along the rod, the wall and the hole go, and the ridge tops out at the nearest rows, 6.8 at y ≈ ±0.067.
- Bend it into a ring, integral(1/hypot(x − cos(u), y − sin(u)), u, 0, 2·π): a charged circle of radius 1, the centre its lowest point inside at 2π ≈ 6.28, ringed by a circular wall.