Electric Flux

We define the electric flux Φ,

of the electric field E,

through the surface A, as:

Where:

A is a vector normal to the surface

(magnitude A, and direction normal to the surface).

θ is the angle between E and A

Φ = E A cos (θ)

Electric Flux

Here the flux is

Φ = E · A

You can think of the flux through some surface as a measure of

the number of field lines which pass through that surface.

Flux depends on the strength of E, on the surface area, and on

the relative orientation of the field and surface.

Normal to surface,

magnitude A

A

E

A ≡

Electric Flux

The flux also depends on orientation

area A

θ

A cos θ

Φ = E . A = E A cos θ

area A

θ

A cos θ

E

E

A

A

Calculate the flux of the electric field E,

through the surface A, in each of the

three cases shown:

​

a) Φ =

b) Φ =

c) Φ =

What if the surface is curved, or the field varies with position ??

1. We divide the surface into small

regions with area dA

2. The flux through dA is

dΦ = E dA cos θ

dΦ = E . dA

​

3. To obtain the total flux we need

to integrate over the surface A

A

Φ = E . A

In the case of a closed surface

The loop means the integral is over a closed surface.

θ

dA

E

For a closed surface:

​

The flux is positive for field lines

that leave the enclosed volume

​

The flux is negative for field lines

that enter the enclosed volume

If a charge is outside a closed surface, the net flux is zero.

As many lines leave the surface, as lines enter it.

For which of these closed surfaces (a, b, c, d)

the flux of the electric field, produced by the

charge +2q, is zero?

Spherical surface with point charge at center

Flux of electric field:

Gauss’s Law

This is always true.

Occasionally, it provides a very easy way

to find the electric field

(for highly symmetric cases).

The electric flux

through any closed surface

equals  enclosed charge / ε₀

Calculate the flux of the electric field Φ

for each of the closed surfaces a, b, c, and d

Surface a, Φ_a =

Surface b, Φ_b =

Surface c, Φ_c =

Surface d, Φ_d =

Calculate the electric field produced

by a point charge using Gauss Law

We choose for the gaussian surface a sphere of radius r, centered on the charge Q.

​

Then, the electric field E, has the same magnitude everywhere on the surface

(radial symmetry)

​

Furthermore, at each point on the surface,

the field E and the surface normal dA are parallel (both point radially outward).

​

E . dA = E dA [cos θ = 1]

E

Q

∫ E . dA = Q / ε₀

∫ E . dA = E ∫ dA = E A

A = 4 π r²

E A = E 4 π r² = Q / ε₀

Electric field produced

by a point charge

E

Q

Is Gauss’s Law more fundamental than Coulomb’s Law?

• No! Here we derived Coulomb’s law for a point charge from Gauss’s law.
• One can instead derive Gauss’s law for a general (even very nasty) charge distribution from Coulomb’s law. The two laws are equivalent.
• Gauss’s law gives us an easy way to solve a few very symmetric problems in electrostatics.
• It also gives us great insight into the electric fields in and on conductors and within voids inside metals.

Gauss’s Law

The total flux within

a closed surface …

… is proportional to

the enclosed charge.

Gauss’s Law is always true, but is only useful for certain

very simple problems with great symmetry.

Applying Gauss’s Law

Gauss’s law is useful only when the electric field

is constant on a given surface

Infinite sheet of charge

A charge Q is uniformly distributed through a sphere of radius R.

What is the electric field as a function of r?. Find E at r₁ and r_2.

Problem: Sphere of Charge Q

A charge Q is uniformly distributed through a sphere of radius R.

What is the electric field as a function of r?. Find E at r₁ and r_2.

Problem: Sphere of Charge Q

Use symmetry!

This is spherically symmetric.

That means that E(r) is radially

outward, and that all points, at a

given radius (|r|=r), have the same

magnitude of field.

Problem: Sphere of Charge Q

E & dA

What is the enclosed charge?

First find E(r) at a point outside the charged sphere. Apply Gauss’s

law, using as the Gaussian surface the sphere of radius r pictured.

Problem: Sphere of Charge Q

r

R

E & dA

What is the enclosed charge? Q

First find E(r) at a point outside the charged sphere. Apply Gauss’s

law, using as the Gaussian surface the sphere of radius r pictured.

Problem: Sphere of Charge Q

r

R

E & dA

What is the enclosed charge? Q

What is the flux through this surface?

First find E(r) at a point outside the charged sphere. Apply Gauss’s

law, using as the Gaussian surface the sphere of radius r pictured.

Problem: Sphere of Charge Q

r

R

E & dA

What is the enclosed charge? Q

What is the flux through this surface?

First find E(r) at a point outside the charged sphere. Apply Gauss’s

law, using as the Gaussian surface the sphere of radius r pictured.

Problem: Sphere of Charge Q

r

R

E & dA

What is the enclosed charge? Q

What is the flux through this surface?

Gauss ⇒

First find E(r) at a point outside the charged sphere. Apply Gauss’s

law, using as the Gaussian surface the sphere of radius r pictured.

Problem: Sphere of Charge Q

r

R

E & dA

What is the enclosed charge? Q

What is the flux through this surface?

Gauss:

So

Exactly as though all the charge were at the origin!

(for r>R)

First find E(r) at a point outside the charged sphere. Apply Gauss’s

law, using as the Gaussian surface the sphere of radius r pictured.

Problem: Sphere of Charge Q

R

r

E(r)

Next find E(r) at a point inside the sphere. Apply Gauss’s law,

using a little sphere of radius r as a Gaussian surface.

What is the enclosed charge?

That takes a little effort. The little sphere has

some fraction of the total charge. What fraction?

That’s given by volume ratio:

Again the flux is:

Setting

gives

For r<R

Problem: Sphere of Charge Q

Problem: Sphere of Charge Q

Look closer at these results. The electric field at comes from a sum over the contributions of all the little bits .

It’s obvious that the net E at this point will be horizontal.

But the magnitude from each bit is different; and it’s completely

not obvious that the magnitude E just depends on the distance

from the sphere’s center to the observation point.

Doing this as a volume integral would be HARD.

Gauss’s law is EASY.

r > R

σ

Problem: Infinite charged plane

Consider an infinite plane with a constant surface charge density σ

(which is some number of Coulombs per square meter).

What is E at a point located a distance z above the plane?

x

y

z

σ

Problem: Infinite charged plane

Consider an infinite plane with a constant surface charge density σ

(which is some number of Coulombs per square meter).

What is E at a point located a distance z above the plane?

x

y

z

Use symmetry!

The electric field must point straight away

from the plane (if σ > 0). Maybe the

Magnitude of E depends on z, but the direction

is fixed. And E is independent of x and y.

E

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

So choose a Gaussian surface that is a “pillbox”, which has its top

above the plane, and its bottom below the plane, each a distance z

from the plane. That way the observation point lies in the top.

z

z

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Total charge enclosed by box =

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Total charge enclosed by box = Aσ

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Outward flux through the top:

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Outward flux through the top: EA

​

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Outward flux through the top: EA

Outward flux through the bottom:

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Outward flux through the top: EA

Outward flux through the bottom: EA

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Outward flux through the top: EA

Outward flux through the bottom: EA

Outward flux through the sides:

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Let the area of the top and bottom be A.

Outward flux through the top: EA

Outward flux through the bottom: EA

Outward flux through the sides: E x (some area) x cos(90⁰) = 0

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Outward flux through the top: EA

Outward flux through the bottom: EA

Outward flux through the sides: E x (some area) x cos(90⁰) = 0

So the total flux is: 2EA

Let the area of the top and bottom be A.

E

E

Gaussian “pillbox”

σ

Problem: Infinite charged plane

z

z

Gauss’s law then says that Aσ/ε₀=2EA so that E=σ/2ε₀, outward.

This is constant everywhere in each half-space!

Let the area of the top and bottom be A.

Notice that the area A canceled: this is typical!

Problem: Infinite charged plane

Imagine doing this with an integral over the charge distribution:

break the surface into little bits dA …

Doing this as a surface integral would be HARD.

Gauss’s law is EASY.

Consider a long cylindrical charge distribution of radius R,

with charge density ρ = a – b r (with a and b positive).

Calculate the electric field for:

1. r < R
2. r = R
3. r > R

• A conductor is a material in which charges can move relatively freely.

​

• Usually these are metals (Au, Cu, Ag, Al).

​

• Excess charges (of the same sign) placed on a conductor will move as far from each other as possible, since they repel each other.

​

• For a charged conductor, in a static situation, all the charge resides at the surface of a conductor.

​

• For a charged conductor, in a static situation, the electric field is zero everywhere inside a conductor, and perpendicular to the surface just outside

​

Conductors

Conductors

Why is E = 0 inside a conductor?

Conductors

Why is E = 0 inside a conductor?

Conductors are full of free electrons, roughly one per

cubic Angstrom. These are free to move. If E is

nonzero in some region, then the electrons there feel

a force -eE and start to move.

Conductors

Why is E = 0 inside a conductor?

Conductors are full of free electrons, roughly one per

cubic Angstrom. These are free to move. If E is

nonzero in some region, then the electrons there feel

a force -eE and start to move.

​

In an electrostatics problem, the electrons adjust their

positions until the force on every electron is zero (or

else it would move!). That means when equilibrium is

reached, E=0 everywhere inside a conductor.

Conductors

Because E = 0 inside, the inside of a conductor is neutral.

Electric field just outside a charged conductor