Electrostatics: Electric field
Continuous charge distributions

1. Electric field due to
continuous charge distributions

A macroscopic objects such as rings, rods, disks, or spheres are charged when they have an imbalance of electrons and protons populations. The charge of such an object is treated as continuous distribution of elementary charges dq.

The electric field due to this charge dq is dE = k dq/r ^r where r is the distance from the element of charge dq to the point P where the electric field is evaluated, and ^r is the unit vector that points from dq to the point P.

The electric field due to the whole object at the point P is E = ∫ dE, or $$\Large\bf\color{green}{ \vec{E} = k\int\frac{dq}{r^2}\hat{r} }$$

2. Examples

2.1. Line charge

A line charge is a distribution of charge spread along long and thin wire. A uniform line charge has a linear charge density λ that is equal to the charge Q of the whole line divided by its length l. λ = Q/l.

We want to find the expression of the electric field E in the perpendicular bisector plane of a uniform line charge.

We have

λ = Q/l = dq/dz

The contribution of dq is dE = k dq/(z² + y²).

The elementary field has two components:
dEy = dE cos θ
dEz = dE sin θ


cos θ = y/[y² + z²]^1/2
sin θ = z/[y² + z²]^1/2 =

Therefore

dEy = k dq/(z² + y²) y/[y² + z²]^1/2 =
k y dq/(y² + z²)^3/2 = k y λ dz/(y² + z²)^3/2

dEz = k dq/(z² + y²) z/[y² + z²]^1/2 =
k z dq/(y² + z²)^3/2 = k z λ dz/(y² + z²)^3/2

The contribution along the y-axis is:

E(y) = ∫ dEy
$$\Large\bf\color{red}{ E(y) = \int\limits_{-l}^{+l} k y \lambda \frac{dz}{(y^2 + z^2)^{\frac{3}{2}}} = k y \lambda \int\limits_{-l}^{+l} \frac{dz}{(y^2 + z^2)^{\frac{3}{2}}} }$$
We have $\Large\bf\color{indigo}{ \int\limits_{-l}^{+l} \frac{dz}{(z^2 + y^2)^{\frac{3}{2}}} = \frac{2l}{y^2 \sqrt {y^2 + l^2}} }$

For this result, go to   ∫ dz/(y² + z²)^3/2 dz



Replacing this result in the expresion of E(y), we find:

E(y) = 2lkλy/y²[y² + l²]^1/2

E(y) = 2lkλ/y(y² + l²)^1/2


The contribution along the z-axis is:

E(z) = ∫ dEz
$$\Large\bf\color{red}{ E(z) = \int\limits_{-l}^{+l} k \lambda z\frac{dz}{(y^2 + z^2)^{\frac{3}{2}}} }$$
The function f(z) = z /(z² + y²)^3/2 is an odd function, so its definite integral between - l and + l is null. That is the contribution of the upper half is cancelled by the contribution of the lower half.

Finally the expression of the electric field E(y) for a line charge of length l and linear charge density λ at a point y in the perpendicular bisector plane is

$$\Large\bf\color{green}{ E(y) = \frac {2 k \lambda l}{y (y^2 + l^2)^{\frac{1}{2}}} }$$

2.2. circular ring

A circular ring contained in the plan xz, centered at the origin, of charge Q, and of radius R, is narrow enough to be considered as circular line charge. As for the above example of the line charge, by symmetry, only the y-components contributes.

dE = k dq cos θ/r²

r² = R² + y²
r is constant for a fixed y.

The complete electric field E at the position y due to the whole ring is:

E = ∫ dE = k cos θ/r² ∫ dq = k Q cos θ/r²

cos θ = y/r = y/(R² + y²)^1/2

Therefore

E(y) = k Q y/(R² + y²)^3/2

$$\Large\bf\color{brown}{ E(y) = \frac {k Q y}{(R^2 + y^2)^{\frac{3}{2}}} }$$

2.3. surface charge

The total charge of the disk is Q = 2πR² σ. σ is the surface charge density of the disk .

The elementary charge the related washer at the position z is dq = σ 2π z dz . Therefore

The elementary electric field at the position y is
dE(y) = k dq cos θ/r² = k σ 2π z dz cos θ/r² =
k σ 2π z dz y/(y² + z²)^3/2

Therefore

The total electric field at the point y due to the hole disk is:

E(y) = ∫ dE

	+ R
E(y) = ∫		k σ 2π z dz y/(y2 + z2)3/2
	0

	+ R
= k σ 2π ∫		z dz y/(y2 + z2)3/2
	0

A primitive of z/(y² + z²)^3/2 is - 1/(y² + z²)^1/2

So

E(y) = k y σ 2π [1/(y²)^1/2 - 1/(y² + R²)^1/2] =
σ y /2ε_o [1/(y²)^1/2 - 1/(y² + R²)^1/2]

E(y) = (σy/2ε_o)[1/(y²)^1/2 - 1/(y² + R²)^1/2]

2.4. cylinder charge

From the formula of the surface charge, we determine the electric field at the point y due to the whole cylinder of depth h.

The electric field for a disk, found above, becomes the elementary electric field for the cylinder.

dE(y:cylinder) = E(y:surface charge) =
(σy/2ε_o)[1/(y²)^1/2 - 1/(y² + R²)^1/2]

Using the relationship:

σ = ρ dy

where ρ is the volume charge density, we obtain:

dE(y:cylinder) = (ρy/2ε_o)[1/(y²)^1/2 - 1/(y² + R²)^1/2 dy

dy goes from yo + h to yo

Therefore

E(y) = ∫ E(y:surface charge)

	yo
E(y) = ∫		(ρy/2εo)[1/(y2)1/2 - 1/(y2 + R2)1/2] dy
	yo + h

	yo
= (ρ/2εo)∫		y[1/(y2)1/2 - 1/(y2 + R2)1/2] dy
	yo + h

We have:

∫ y[1/(y²)^1/2 dy = y
∫ - y/(y² + R²)^1/2 dy = - (y² + R²)²


Therefore

	yo
E(y) = (ρ/2εo)∫		y[1/(y2)1/2 - 1/(y2 + R2)1/2] dy
	yo + h

= (ρ/2ε_o)[- h + [(yo + h)² + R²]² - [yo² + R²]²]

E(y) = (ρ/2ε_o)[- h + [(yo + h)² + R²]² - (yo² + R²)²]