Definitions:
Source charge: charges of which the positions are given (as functions of time)
Test charge: charges that are introduced and of which the trajectory needs to be calculated.
Principle of Superposition: The idea that the interaction between two things are completely unaffected by others, and thus,
can be individually computed and then added together.
Equipotential: A surface over which the potential is constant.
1.5: Dirac Delta Function
Delta function is zero everywhere, except when its argument is zero, then it will go to infinity. Also, the area of the function is 1
(integrated over +- infinity will give 1). So, the product of f(x) with the delta function in an integral, will result in just the f(x)
(as a constant) with x as the value that yields infinity for the delta function. The delta function itself is not a legitimate function,
but an integral over it, is. Z ∞
f (x)δ(x − a)dx = f (a)
−∞
This can be dragged over 3D, where the delta function will be written as δ 3 .
2.1: The Electric Field
Coulomb’s Law
Force on a test charge Q due to a single point charge q, that is at rest a distance r away:
1 qQ
F = r̂
4πϵ0 r2
r is a separation vector (r = r - r’), where r is the normal distance to the spot, and r’ is the distance to the coordinate point.
Similar sign in this equation, means repelling force (a larger than 0 if F=ma).
By adding more point charges q, we can see that the force can be separated as F = QE. E varies from point to point, which is
why it is r̂-dependent.
Electric Field
For continuous charge distributions, the formula changes from a sum to an integral:
n
λ(r′ ) (r − r′ )
Z
1 X qi 1 ′ r
E(r) = 2 r̂i → E(r) = 2
r̂ dl , where r̂ is calculated as: = (1)
4πϵ0 i=1 ri 4πϵ0 r |r| |r|
NOTE: unit vector r is not a constant, since its direction depends on the source point r’. Only Cartesian coordinates x, y, and z
can be taken out of the integral since these as unit vectors are independent.
2.2: Divergence and Curl of Electrostatic Fields
Fields & its lines
• Falls off like 1/r2 : vectors get shorter the further away
• Always points radially outwards from source
• Magnitude of the field is indicated by density of the field lines (in 3D, also adheres the exponential decrease when going
radially outward)
• Number of lines needs to be proportional to the charge of the particle.
• Go from positive to negative charges (cannot terminate mid-air: Divergence would not be zero)
Flux
The flux through a certain surface is the surface integral of E over that surface. Dot product in the equation picks out just the
part of the surface that aligns with E: perpendicular to the direction of the field lines.
Flux through any closed surface means that is a measure of the enclosed charge, for field lines that come from the charge
need to either exit the surface, or return to an oppositely charged particle within the surface. Outside of the surface will contribute
nothing, since all field lines will eventually return back. For any closed surface:
, Gauss’s Law
I Z Z
1 1
E · da = (∇ · E)dτ = Qencl = ρ dτ (2)
V ϵ0 ϵ0 V
by application of the divergence theorem and a definition of charge. From this follows that:
1
∇·E = ρ (3)
ϵ0
For the volume integrals, it doesn’t matter what volume you take, since the density will be zero outside of the object, so it won’t
add anything.
Gauss’s Law only works with symmetry, since otherwise the electric field will not point in the same direction as da, and its
magnitude would not be constant (due to the dot product). This is why you must use either spherical symmetry, cylindrical
symmetry or plane symmetry.
Curl
I
E · dl = 0 → ∇ × E = 0
2.3: Electric Potential
Because the electric field is path-independent, we can define a potential, for which E = −∇V , and thus:
Z r
E · dl = V (a) − V (b) (4)
0
according to the divergence theorem. Only possible if path-independent, otherwise the path of E would influence the value of V,
and thus, this value could not be defined to be a single function.
Changing reference points will not influence the difference in potential. Rather, it would give another constant K to the result
of the integral, as a result from the line integral of E between the old and new reference point. It does not, of course, change E.
Potential, just like the electrical field and the forces, obey the superposition principle.
Units: force is Newtons, charge is coulombs, Electric fields are Newtons per coulomb, potential is newton-meters per coulomb/joules
per coulomb, which is VOLT.
Potential within a sphere, for instance, where there isn’t any charge, results in a constant potential.
Poisson and Laplace
Using the definition of the curl and divergence of E, we can get that the divergence of E is the Laplacian of V, or then with Gauss’s
law:
ρ
∇2 V = − or, when there is no charge,: ∇2 V = 0 (5)
ϵ0
For continuous charge distribution, the potential is shown as follows:
Z
1 q 1 ρ(r) ′
V (r) = → V (r) = dτ (6)
4πϵ0 r 4πϵ0 r
This is similar to an electric field equation, but there is no unit vector r anymore, and the radius is no long squared. This formula
assumes that the origin is infinity, so it falls apart when the charge itself extends to infinity.
Boundary conditions
1
When crossing a charged surface, the electrical field needs to change with ϵ0 ∗ Qencl . This means that when crossing a boundary,
the orthogonal component of an electric field is discontinuous by:
E ⃗ below = σ n̂ → ∇Vabove−below = − σ n̂
⃗ above − E (7)
ϵ0 ϵ0
Of which the latter is the normal derivative of V.
2.4: Work and Energy in Electrostatics
Z b Z b
W = F · dl = −Q E · dl = Q[V (b) − V (a)] and with reference at infinity: W = QV (⃗r) (8)
a a
In order to compute the collection of charges to a certain place, you use:
n n n
1 X X qi qj 1X
W = →W = qi V (⃗
ri ) (9)
4πϵ0 i=1 j>1 rij 2 i=1
It represents the energy stored in a system, which you will get back by dismantling or need when constructing.
Source charge: charges of which the positions are given (as functions of time)
Test charge: charges that are introduced and of which the trajectory needs to be calculated.
Principle of Superposition: The idea that the interaction between two things are completely unaffected by others, and thus,
can be individually computed and then added together.
Equipotential: A surface over which the potential is constant.
1.5: Dirac Delta Function
Delta function is zero everywhere, except when its argument is zero, then it will go to infinity. Also, the area of the function is 1
(integrated over +- infinity will give 1). So, the product of f(x) with the delta function in an integral, will result in just the f(x)
(as a constant) with x as the value that yields infinity for the delta function. The delta function itself is not a legitimate function,
but an integral over it, is. Z ∞
f (x)δ(x − a)dx = f (a)
−∞
This can be dragged over 3D, where the delta function will be written as δ 3 .
2.1: The Electric Field
Coulomb’s Law
Force on a test charge Q due to a single point charge q, that is at rest a distance r away:
1 qQ
F = r̂
4πϵ0 r2
r is a separation vector (r = r - r’), where r is the normal distance to the spot, and r’ is the distance to the coordinate point.
Similar sign in this equation, means repelling force (a larger than 0 if F=ma).
By adding more point charges q, we can see that the force can be separated as F = QE. E varies from point to point, which is
why it is r̂-dependent.
Electric Field
For continuous charge distributions, the formula changes from a sum to an integral:
n
λ(r′ ) (r − r′ )
Z
1 X qi 1 ′ r
E(r) = 2 r̂i → E(r) = 2
r̂ dl , where r̂ is calculated as: = (1)
4πϵ0 i=1 ri 4πϵ0 r |r| |r|
NOTE: unit vector r is not a constant, since its direction depends on the source point r’. Only Cartesian coordinates x, y, and z
can be taken out of the integral since these as unit vectors are independent.
2.2: Divergence and Curl of Electrostatic Fields
Fields & its lines
• Falls off like 1/r2 : vectors get shorter the further away
• Always points radially outwards from source
• Magnitude of the field is indicated by density of the field lines (in 3D, also adheres the exponential decrease when going
radially outward)
• Number of lines needs to be proportional to the charge of the particle.
• Go from positive to negative charges (cannot terminate mid-air: Divergence would not be zero)
Flux
The flux through a certain surface is the surface integral of E over that surface. Dot product in the equation picks out just the
part of the surface that aligns with E: perpendicular to the direction of the field lines.
Flux through any closed surface means that is a measure of the enclosed charge, for field lines that come from the charge
need to either exit the surface, or return to an oppositely charged particle within the surface. Outside of the surface will contribute
nothing, since all field lines will eventually return back. For any closed surface:
, Gauss’s Law
I Z Z
1 1
E · da = (∇ · E)dτ = Qencl = ρ dτ (2)
V ϵ0 ϵ0 V
by application of the divergence theorem and a definition of charge. From this follows that:
1
∇·E = ρ (3)
ϵ0
For the volume integrals, it doesn’t matter what volume you take, since the density will be zero outside of the object, so it won’t
add anything.
Gauss’s Law only works with symmetry, since otherwise the electric field will not point in the same direction as da, and its
magnitude would not be constant (due to the dot product). This is why you must use either spherical symmetry, cylindrical
symmetry or plane symmetry.
Curl
I
E · dl = 0 → ∇ × E = 0
2.3: Electric Potential
Because the electric field is path-independent, we can define a potential, for which E = −∇V , and thus:
Z r
E · dl = V (a) − V (b) (4)
0
according to the divergence theorem. Only possible if path-independent, otherwise the path of E would influence the value of V,
and thus, this value could not be defined to be a single function.
Changing reference points will not influence the difference in potential. Rather, it would give another constant K to the result
of the integral, as a result from the line integral of E between the old and new reference point. It does not, of course, change E.
Potential, just like the electrical field and the forces, obey the superposition principle.
Units: force is Newtons, charge is coulombs, Electric fields are Newtons per coulomb, potential is newton-meters per coulomb/joules
per coulomb, which is VOLT.
Potential within a sphere, for instance, where there isn’t any charge, results in a constant potential.
Poisson and Laplace
Using the definition of the curl and divergence of E, we can get that the divergence of E is the Laplacian of V, or then with Gauss’s
law:
ρ
∇2 V = − or, when there is no charge,: ∇2 V = 0 (5)
ϵ0
For continuous charge distribution, the potential is shown as follows:
Z
1 q 1 ρ(r) ′
V (r) = → V (r) = dτ (6)
4πϵ0 r 4πϵ0 r
This is similar to an electric field equation, but there is no unit vector r anymore, and the radius is no long squared. This formula
assumes that the origin is infinity, so it falls apart when the charge itself extends to infinity.
Boundary conditions
1
When crossing a charged surface, the electrical field needs to change with ϵ0 ∗ Qencl . This means that when crossing a boundary,
the orthogonal component of an electric field is discontinuous by:
E ⃗ below = σ n̂ → ∇Vabove−below = − σ n̂
⃗ above − E (7)
ϵ0 ϵ0
Of which the latter is the normal derivative of V.
2.4: Work and Energy in Electrostatics
Z b Z b
W = F · dl = −Q E · dl = Q[V (b) − V (a)] and with reference at infinity: W = QV (⃗r) (8)
a a
In order to compute the collection of charges to a certain place, you use:
n n n
1 X X qi qj 1X
W = →W = qi V (⃗
ri ) (9)
4πϵ0 i=1 j>1 rij 2 i=1
It represents the energy stored in a system, which you will get back by dismantling or need when constructing.
