Electric Field of a Charged Disk

1 Electric Field of a Charged Disk

The Electric Field of a charged disk can teach us important concepts that you will see over and over in physics: superposition, cylindrical coordinates and non-constant basis vectors. To get a glimpse of the power of superposition, we will solve this problem the hard way first and then see how superposition can be a powerfull tool.

Let us calculate the Electric Field at a point P above the center of a charged disk with radius of R and a uniform surface charge density of σ as shown in below figure.

Starting with the general formula for a surface charge

∫ 1 σ (r′)(r − r′)da ′ E = 4π𝜖-- ----------′3--- 0 |r − r|

(1)

choose a coordinate system. A disk clearly lends itself to cylindrical coordinates. As a refresher, the next figure shows the infinitesimal displacement, where we have the infinitesmal area da′

cartesian coordinates:

da′ = dx′dy′

cylindrical coordinates:

da′ = s′ds ′dϕ ′

The vectors to the source and field points that are needed for the integration in cylindrical coordinates

r = z ˆz

′ ′ˆ′ r = s ϕ

therefore

r − r′ = zˆz − s′ϕˆ′

√ -------- |r − r′| = s′2 + z2

substituting these relationships into (1) gives us

σ ∫ 2π∫ R s′ds′dϕ′ ( ) E = ----- ---------3∕2 zˆz − s′ϕˆ′ 4π𝜖0 0 0 (s′2 + z2)

(2)

As usual break up the integration into the z and ϕ components

z component:

σ ∫ 2π∫ R z ˆzs′ds ′dϕ ′ Ez = ----- -′2----23∕2- 4π𝜖0 0 0 (s + z )

Since ẑ is always in the same direction and has the same magnitude (unit vector), it is constant and can be brought out of the integration. Integrating the ds them

∫ ∫ σz-ˆz- 2π ′ R ----s′ds-′--- Ez = 4π 𝜖0 dϕ ′2 2 3∕2 0 0 (s + z )

using u substitution

′2 2 u = s + z

du = 2s′ds

du- ds = 2s

with the limits of integration becoming

u(s′ = 0 ) = z2

u(s′ = R ) = R2 + z2

trasnforming the integral to

∫ 2π ∫ R2+z2 − 3∕2 Ez = -σzˆz- dϕ′ u----du- 4π 𝜖0 0 z2 2

integrating

∫ 2π σz-ˆz- ′ R2+z2 −1∕2 Ez = 4π 𝜖0 0 d ϕ |z2 − u du

evaluating the limits

∫ 2π ( ) -σzˆz- 1- ----1----- ′ Ez = 4π 𝜖0 0 z − √ R2 + z2 dϕ

integrating again simply gives

( ) σzˆz 1 1 Ez = ---- --− √---2----2 2𝜖0 z R + z

ϕ component:

∫ ∫ -σ--- 2π R -s′2ˆϕ′ds-′dϕ-′ E ϕ = 4π 𝜖 − ′2 23∕2 0 0 0 (s + z )

If you cannot simply see how the ϕ component is zero through symmetry, then carry out the integration. The key thing to learn here, and why it is not good to just skip over the ϕ component, is to realize that ϕ is not constant throughout the integration. Therefore, one cannot bring it out of the integration. What needs to be done is to substitute in for ϕ. An important result from cylindrical coordinates is the relation between its unit vectros and those of cartesian coordinates.