In this page, we are going to calculate the **electric field due to a thin disk of charge**. We will assume that the charge is homogeneously distributed, and therefore that the surface charge density σ is constant. We will also assume that the total charge q of the disk is positive; if it were negative, the electric field would have the same magnitude but an opposite direction.

As we will see later, the electric field due to an infinite thin sheet of charge is a particular case of the field due to a thin disk of charge.

##### Electric field due to a ring of charge

As a previous step we will calculate the **electric field due to a ring of positive charge** at a point P located on its axis of symmetry at a distance *x* of the ring (see next figure). The total charge of the ring is *q * and its radius is *R’*.

We will first calculate the electric field due to a charge element *dq* (in red in the figure) located at a distance *r* from point P. The charge element can be considered as a point charge, thus the electric field due to it at point P is:

And the total electric field due to the ring is the following integral:

Before evaluating this type of integrals, it is convenient to first analyze the symmetry of the problem to see if it can be simplified.

### Ad blocker detected

Both the electric field *d E* due to a charge element

*dq*and to another element with the same charge but located at the opposite side of the ring is represented in the following figure. The ring is positively charged so

*dq*is a

**source of field lines**, therefore d

**E**is directed outwards. Furthermore, the electric field satisfies the superposition principle, so the total electric field at point P is the sum of the individual electric fields due to both charge elements:

As you can see in the previous figure, the vertical component of the vector sum of both fields *d E* is zero. This is true for any charge element and their diametrically opposed; therefore, the magnitude of the total electric field is the integral of the horizontal projections of

*d*. The magnitude of the electric field due to the ring at point P is therefore:

**E**Where the integral is taken over the whole ring.

As seen in the figure, the cosine of angle α and the distance r are respectively:

And after substituting we obtain:

This expression will allow us to calculate the electric field due to a thin disk of charge.

Finally, we integrate to calculate the field due to a ring of charge at point P:

### Ad blocker detected

##### Electric field due to a thin disk of charge

We will calculate the electric field due to the thin disk of radius R represented in the next figure. The total charge of the disk is *q*, and its surface charge density is σ (we will assume it is constant). We will use a ring with a radius *R’* and a width *dR’* as charge element to calculate the electric field due to the disk at a point P located on its axis of symmetry.

The electric field *dE*_{x} due to the charge element is similar to the electric field due to a ring calculated before:

We have to integrate the previous expression over the whole charge distribution to calculate the total field due to a disk. We have to express *dq* in such a way that we can solve the integral and to do so we will use the definition of the surface charge density:

Where 2πR’dR’ is the surface area of the circular ring represented in the previous figure.

After substituting in the expression of the electric field *dE*_{x} and simplifying we obtain:

Finally, after solving the integral we get:

##### Electric field due to infinite thin sheet of charge

An infinite thin sheet of charge is a particular case of a disk when the radius R of the disk tends to infinity (R → ∞)

The limit of the electric field due to a disk when R → ∞ is:

You can see how to calculate the magnitude of the electric field due to an infinite thin sheet of charge using Gauss’s law in this page.

The post Electric field due to a ring, a disk and an infinite sheet appeared first on YouPhysics