2Electrostatics

IB Electromagnetism

2.1 Gauss’ Law

Here we transform the first Maxwell’s equation into an integral form, known as

Gauss’ Law.

Consider a region

V ⊆ R

3

with boundary

S

=

∂V

. Then integrating the first

equation over the volume V gives

Z

V

∇ · E dV =

1

ε

0

Z

V

ρ dV.

The divergence theorem gives

R

V

∇ · E

d

V

=

R

S

E ·

d

S

, and by definition,

Q =

R

V

ρ dV . So we end up with

Law (Gauss’ law).

Z

S

E · dS =

Q

ε

0

,

where Q is the total charge inside V .

Definition

(Flux through surface)

.

The flux of

E

through the surface

S

is

defined to be

Z

S

E · dS.

Gauss’ law tells us that the flux depends only on the total charge contained

inside the surface. In particular any external charge does not contribute to the

total flux. While external charges do create fields that pass through the surface,

the fields have to enter the volume through one side of the surface and leave

through the other. Gauss’ law tells us that these two cancel each other out

exactly, and the total flux caused by external charges is zero.

From this, we can prove Coulomb’s law:

Example

(Coulomb’s law)

.

We consider a spherically symmetric charge density

ρ

(

r

) with

ρ

(

r

) = 0 for

r > R

, i.e. all the charge is contained in a ball of radius

R

.

R

E

S

By symmetry, the force is the same in all directions and point outward radially.

So

E = E(r)

ˆ

r.

This immediately ensures that ∇ × E = 0.

Put S to be a sphere of radius r > R. Then the total flux is

Z

S

E · dS =

Z

S

E(r)

ˆ

r · dS

= E(r)

Z

S

ˆ

r · dS

= E(r) · 4πr

2

By Gauss’ law, we know that this is equal to

Q

ε

0

. Therefore

E(r) =

Q

4πε

0

r

2

and

E(r) =

Q

4πε

0

r

2

ˆ

r.

By the Lorentz force law, the force experienced by a second charge is

F(r) =

Qq

4πε

0

r

2

ˆ

r,

which is Coulomb’s law.

Strictly speaking, this only holds when the charges are not moving. However,

for most practical purposes, we can still use this because the corrections required

when they are moving are tiny.

Example. Consider a uniform sphere with

ρ(r) =

(

ρ r < R

0 r > R

.

Outside, we know that

E(r) =

Q

4πε

0

r

2

ˆ

r

Now suppose we are inside the sphere.

R

E

r

Then

Z

S

E · dS = E(r)4πr

2

=

Q

ε

0

r

3

R

3

So

E(r) =

Qr

4πε

0

R

3

ˆ

r,

and the field increases with radius.

Example (Line charge). Consider an infinite line with uniform charge density

per unit length η.

We use cylindrical polar coordinates:

z

r =

p

x

2

+ y

2

E

By symmetry, the field is radial, i.e.

E(r) = E(r)

ˆ

r.

Pick

S

to be a cylinder of length

L

and radius

r

. We know that the end caps do

not contribute to the flux since the field lines are perpendicular to the normal.

Also, the curved surface has area 2πrL. Then

Z

S

E · dS = E(r)2πrL =

ηL

ε

0

.

So

E(r) =

η

2πε

0

r

ˆ

r.

Note that the field varies as 1

/r

, not 1

/r

2

. Intuitively, this is because we have

one more dimension of “stuff” compared to the point charge, so the field does

not drop as fast.

Example

(Surface charge)

.

Consider an infinite plane

z

= 0, with uniform

charge per unit area σ.

By symmetry, the field points vertically, and the field bottom is the opposite of

that on top. we must have

E = E(z)

ˆ

z

with

E(z) = −E(−z).

Consider a vertical cylinder of height 2

z

and cross-sectional area

A

. Now only

the end caps contribute. So

Z

S

E · dS = E(z)A − E(−z)A =

σA

ε

0

.

So

E(z) =

σ

2ε

0

and is constant.

Note that the electric field is discontinuous across the surface. We have

E(z → 0+) − E(z → 0−) =

σ

ε

0

.

This is a general result that is true for any arbitrary surfaces and

σ

. We can prove

this by considering a cylinder across the surface and then shrink it indefinitely.

Then we find that

ˆ

n · E

+

−

ˆ

n · E

−

=

σ

ε

0

.

However, the components of E tangential to the surface are continuous.