Electric Field Due to a Charge Distribution

We consider a volume (or line or area, as appropriate) element of the object. We set an origin and determine the location of the field point P and source point S in rectilinear coordinates.

The electric field due to the element is given by

Note that a = P - S. Express a in rectilinear coordinates.

The element chosen should reflect the symmetry of the object - linear, rectilinear, polar, cylindrical, or spherical. This makes the limits of integration easier to find.

The angle α will occur in your solution since dE_x = dEcos(α) and dE_y = dEsin(α). Recall that

and

By doing the dot and cross products, cos(α) and sin(α) can be expressed in simpler form.

The charge element dq is related to the element of integration by the charge density

where λ, Σ, and ρ are the linear, surface, and volume charge densities respectively.

Coordinate System	Elements	ds (linear)	dA (area)	dV (volume)
rectilinear (lines, rectangles, cubes)	dx, dx, dy	dx	dxdy	dxdydz
polar (arcs, circles, pie slices)	dr, rdθ	Rdθ	rdrdθ	-
cylindrical polar (cans, cylinders)	dz, dr, rdθ	-	Rdθdz	rdrdθdz
spherical (balls, part of spheres)	dr, rdθ, rsin(θ)dφ	-	R2sin(θ) dθdφ	r2sin(θ) dθdφdr

Conversion to rectilinear coordinates

Coordinate System	x	y	z	Range
polar	rsin(θ)	rcos(θ)	-	0 £ q £ 2π	-
cylindrical polar	rsin(θ)	rcos(θ)	z	0 £ q £ 2π	-
spherical	rsin(θ)cos(θ)	rsin(θ)sin(θ)	rcos(θ)	0 £ q £ π	0 £ f £ 2π

Example

What is the electric field at point P = (X, 0, 0) due to a disk of radius R and charge density located at the origin? Note that X > R.

The components of the electric field are dE_x = dE cos(α) and dE_y = -dE sin(α), where

Now dq = rdrdθ. As well, S = (rcos(θ), rsin(θ)). Thus the vector a is

We also need the magnitude of a,

The cosine and sine are

and

Our integrals, therefore, are

and

Questions? mike.coombes@kwantlen.ca