EM-04 · Electrostatics field

Integrating Continuous Charge

The electric field of charge spread out continuously is found by splitting it into many point-charge pieces and adding up, as vectors, the small field each piece makes. You see for yourself that the finer the pieces, the closer the sum gets to the true value.

Add pieces and close in on the integral

Choose with the buttons how many pieces to cut the charged rod into, and the small arrows dE from each piece link head to tail to form the total electric field E at point P. With only one piece the sum misses even in direction, but as the pieces increase, the end of the chain settles on one point. The convergence on the indicator shows, as a percentage, how close the sum is to the true value.

Number of pieces N

Drag to rotate; the buttons change the number of pieces (N).

The field E at P (sum of the pieces)

E = ∫ dE = ∫ k dq/r² r̂ · sum of tiny fields

dq = λ dl = σ dA = ρ dV · line, surface, volume density

Convergence · 70%

Coarse

From a point to a distribution

You saw the electric field of a single point charge in EM-03. When charge is spread continuously, like a rod or a plate, you can treat it as many point-charge pieces. The small field made by one piece is dE = k dq/r² r̂. The total field is the sum of all these small fields, and that sum taken over infinitely fine pieces is an integral.

ObservedE = k dqr² r̂

One piece makes a dE just like a point charge.

The charge of a piece, dq

The charge dq of one piece is the charge density times the size of the piece. For a line charge dq = λ dl, for a surface charge dq = σ dA, and for a volume charge dq = ρ dV. Here dl, dA, dV are the small elements chosen in EM-01, so picking a coordinate system that fits the symmetry keeps this integral clean too.

Choosedq = ?

The charge of a line-charge piece is the line density times its length.

The sum becomes an integral

Cut the rod into N pieces and the vector sum of N small fields is an approximation. Grow N to make the pieces infinitely small, and the sum settles on one value; that limit is the integral E = ∫ k dq/r² r̂. Since the electric field is a vector, in practice each component is integrated separately. For the rod in the figure, cutting it into just four pieces already brings the convergence close to one hundred percent, because the smaller the pieces, the better each one is approximated as a point charge.

Fill inE = ?

An integral adds up the small fields over infinitely fine pieces.

On your ownE = ?

The total field is the integral of dE over every piece.

Directions add up too

The small fields dE are vectors, so direction adds along with magnitude. Where there is symmetry, some components cancel. Beside the middle of a very long line charge, the components along the line all cancel and only the perpendicular component remains, which makes the answer simple. Point P in the figure lies beyond the end of the rod, so there is no such cancellation and components in several directions remain. Before starting an integral, the trick is to check first which components survive.

Back to the first screen

On the first screen, with one piece the rod was treated as a single point in the middle, so the arrow missed even in direction. As the pieces increased, the small fields dE linked head to tail and the sum settled on one direction and one size. That converged value is the integral E = ∫ k dq/r² r̂. In the end, an integral is the job of adding up, without missing any, the fields of finely cut point charges.

The electric field of a continuous charge is found by splitting the distribution into countless point-charge pieces dq and integrating as vectors the small fields dE = k dq/r² r̂ of each piece. E = ∫ k dq/r² r̂, and the charge of a piece is dq = λ dl (line), σ dA (surface), ρ dV (volume).

What comes next

Direct integration works for any distribution but takes a lot of work. For distributions with good symmetry there is a faster road. Go on to flux and divergence (EM-05), which count how much electric field passes through a closed surface, and to Gauss's law (EM-06), and you can find the field of a symmetric distribution without integrating.