The fundamental equations of electrostatics are **linear equations**,

**∇**∙**E **= ρ/ε_{0}, **∇**×**E** = 0, (SI units).

The **principle of superposition** holds.

The** electrostatic force **on a particle with charge q at position **r**
is **F **= q**E**(**r**).

**∇**×**E **= 0 <==> **E** = -**∇**Φ, ∇^{2}Φ =
-ρ/ε_{0}.

Φ is the **electrostatic potential**.

The **field** at r due to a point charge at r': **E**(**r**)
= [1/(4πε_{0})] q(**r** - **r**')/|**r**
- **r**'|^{3}.

The field of a charge distribution:

**E**(**r**) = [1/(4πε_{0})][∫_{v'}
dV' ρ(**r'**)(**r** - **r**')/|**r** - **r**'|^{3} +
∫_{A' }
dA' σ(**r'**)(**r** - **r**')/|**r** - **r**'|^{3}

+ ∫_{l' }dl' λ(**r'**)(**r** - **r**')/|**r** – **
r**'|^{3} + ∑_{i}q_{i}(**r** - **r**')/|**r**
- **r**_{i}|^{3}].

(We consider volume, surface, and line charge distributions and point
charges.)

The **potential** at **r** due to a point charge at **r**': Φ(**r**) = [1/(4πε_{0})]_{
}q/|**r** - **r**'|.

The potential of a charge distribution:

Φ(**r**) = [1/(4πε_{0})][∫_{v'} dV' ρ(**r'**)/|**r**
- **r**'| + ∫_{A' }dA' σ(**r'**)/|**r** - **
r**'|

+ ∫_{l' }dl' λ(**r'**)/|**r** – **
r**'| + ∑_{i}q_{i}/|**r** - **r**_{i}|].

**Gauss' law: ** ∫_{closed surface} **E·**d**A **= Q_{inside}/ε_{0 }
.

In situations with enough symmetry, Gauss' law alone can be used to find E.

The **electrostatic energy** of a charge distribution:

U = ½∫Φ(**r**)dq(**r**) = ½[∫_{v}dV ρ(**r**)Φ(**r**)
+ ∫_{A}dA σ(**r**)Φ(**r**) + ∫_{l }dl λ(**r**)Φ(**r**)
+ ∑_{i}q_{i}Φ(**r _{i}**)],

or, for a continuous charge distribution, U = (ε

The field of a dipole at the origin: **E**(**r**) = [1/(4πε_{0})](1/r^{3})[3(**p**∙**r**)**r**/r^{2} - **p**].

The potential of a dipole at the origin: Φ(**r**) = [1/(4πε_{0})](**p**∙**r**)/r^{3}.

The force on a dipole: **F** = **∇**(**p**∙**E**).

The torque on a dipole: **τ** = **p**×**E.
**The energy of a dipole in an external field: U = -(

Properties of conductors in electrostatics

**E**= 0 inside.- ρ = 0 inside.
- Any excess charge resides on the surface.
**E**on the surface is perpendicular to the surface.-
**E**= (σ/ε_{0})**n**just outside the surface.

(**E**_{2} - **E**_{1})∙**n**_{2}** **= (σ/ε_{0}),

(**E**_{2} - **E**_{1})∙**t **= 0,

(**D**_{2} - **D**_{1})∙**n**_{2}** **= σ_{f},

Φ is continuous across the boundary.

**The multipole expansion**

For a charge distribution located near the origin we have

Φ(**r**) = [1/(4πε_{0})][(1/r)[∫_{v'}
dV' ρ(**r'**) + (1/r^{2} )(**r**/r)∙[∫_{v'} dV' ρ(**r'**)**r**'

+ (1/r^{3} )[∫_{v'} dV'ρ(**r'**)(3(**r**∙**r**')^{2}/r^{2}
- **r**'^{2})/2 + ... ]

= 1/(4πε_{0})][Q/r)
+ **p**∙(**r**/r)/r^{2}
+ (1/(2r^{5}))∑_{ij}3x_{i}x_{j}Q_{ij} +
... ],

with **p** = ∫_{v'} dV' ρ(**r'**)**r**'
and

Q_{ij} = ∫(x_{i}'x_{j}' - ⅓δ_{ij}r'^{2})
ρ(**r'**)dV', Q_{ij} = Q_{ji}, ∑_{i}Q_{ii} =
0.

Here Q = total charge, **p** = dipole moment, Q_{ij} = quadrupole
moment tensor of the charge distribution. If the problem has rotational
symmetry about the z-axis, such that Q_{xx }= Q_{yy }= -½Q_{zz},
then Q_{zz} is called** **the** quadrupole moment**.

The energy of a charge distribution near the origin in an external field is
given by

W = qΦ(0) - **p**∙**E**(0) - ½∑_{i}∑_{j}Q_{ij}
∂E_{j}/∂x_{i}|_{xi=0} + ... .

This expansion shows how the various multipoles interact with the external
field.

Consider a charge q at z = d on the z - axis.

- Assume the z = 0 plane is a
**grounded conducting plane**. Then placing an image charge q' = -q at d' = -d on the z-axis makes the z = 0 plane an equipotential with Φ = 0. - Assume a
**grounded conducting sphere of radius**R < d is centered on the origin. Then placing an image charge q' = -qR/d at d' = R^{2}/d on the z-axis makes the sphere an equipotential with Φ = 0. Add q'' at the center of the sphere to change the potential or the total charge on the sphere.

Consider a line charge λ parallel to the x-axis at **d** = d **k**.

- Assume a
**conducting cylinder of radius**R < d has as its axis the x-axis. Then placing an image line charge λ' = -λ at d'**k**with d' = R^{2}/d parallel to the x-axis makes the cylinder an equipotential with Φ = [λ/(2πε_{0})]ln(R/d). Add V_{0}everywhere to change the potential of the cylinder.

When solving electrostatic problems, we often rely on the **uniqueness
theorem**. If the charge density ρ is specified throughout a volume V,
and Φ** **or its normal derivatives are specified at the boundaries of a
volume V, then a unique solution exists for Φ inside V.

In regions with ρ = 0 we have
∇^{2}Φ = 0. We are looking for a solution for ∇^{2}Φ
= 0, subject to boundary conditions.

Let
Φ = Φ(r,θ), 0 ≤ θ ≤ π
in **spherical coordinate**s, i.e. let the problem have azimuthal symmetry.

Then the
most general solution for ∇^{2}Φ = 0 is

Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}
+ B_{n}/r^{n+1}]P_{n}(cosθ).

Let
Φ = Φ(ρ,φ), 0 ≤ θ ≤ 2π in
**cylindrical coordinates**.

Then the most general solution for ∇^{2}Φ =
0 is

Φ(ρ,φ) =∑_{n=1}^{∞}(A_{n}cos(nφ)
+ B_{n}sin(nφ))ρ^{n} + ∑_{n=1}^{∞}(A'_{n}cos(nφ)
+ B'_{n}sin(nφ))ρ^{-n} + a_{0} + b_{0}lnρ.