EM-04 · Electrostatics field

Integrating Continuous Charge

The field of a continuous charge is found by chopping it into countless point-charge pieces and adding each piece’s tiny field as a vector. See for yourself that the finer the pieces, the closer the sum gets to the true value.

Add pieces to approach the integral

Choose how many pieces to cut the charged rod into. Joining each piece’s tiny arrow dE head to tail gives the whole field E. As the pieces multiply, how does the sum change?

Number of pieces N

Drag to orbit. Buttons set 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

We saw the field of a single point charge in EM-03. When charge is spread continuously, like a rod or a plate, we treat it as countless point-charge pieces. Each piece dq makes a tiny field dE = k dq/r² r̂. The whole field is the sum of all these tiny fields, and that sum, taken over ever finer pieces, is an integral.

ObservedE = k dqr² r̂

One piece makes a dE just like a point charge.

The charge of a piece, dq

A piece’s charge dq is its density times its size. For a line charge dq = λ dl, for a surface charge dq = σ dA, for a volume charge dq = ρ dV. Here dl, dA, dV are the differential elements you chose back in EM-01. Picking the coordinate system that matches the symmetry keeps this integral clean.

Choosedq = ?

A line piece’s charge is density times length.

The sum becomes an integral

Cut into N pieces, the vector sum of N tiny fields is an approximation. As N grows and the pieces become infinitely fine, the sum converges to a single value. That limit is the integral E = ∫ k dq/r² r̂. Because the field is a vector, you integrate it component by component.

Fill inE = ?

The integral is the infinite sum of tiny fields.

On your ownE = ?

The whole field is the integral of every piece’s dE.

Directions add up too

Because dE is a vector, not just magnitudes but directions add. When there is symmetry, some components cancel. For a very long straight line of charge, for instance, the components along the rod all wash out and only the perpendicular ones remain, so the answer simplifies. Seeing which components survive first is the knack of the integral.

Back to the first screen

On the first screen, with a single piece the rod was treated as one central point, so the arrow missed. As you added pieces, the tiny fields dE joined head to tail and the sum converged to one direction and one magnitude. That converged value is the integral E = ∫ k dq/r² r̂. Integration is, in the end, adding up the fields of finely chopped point charges without leaving any out.

The field of a continuous charge is found by chopping the distribution into countless point-charge pieces dq and integrating each piece’s tiny field dE = k dq/r² r̂ as a vector. So E = ∫ k dq/r² r̂, where the piece charge is dq = λ dl (line), σ dA (surface), or ρ dV (volume).

What comes next

Direct integration works for any distribution but takes effort. When the symmetry is good, there is a faster road. Counting how much field pierces a closed surface — flux and divergence (EM-05) and the Gauss law (EM-06) — lets you get the field of a symmetric distribution without integrating.