Griffiths Ch. 2.1, part of 2.2
Coulomb's law For a single, static point charge
at position , the force exerted on a charge at position is: \vec{F} = \frac{1}{4\pi \epsilon_0} \frac{qQ}{\zeta^2} \hat{\zeta}
F ⃗ = 1 4 π ϵ 0 q Q ζ 2 ζ ^ \vec{F} = \frac{1}{4\pi \epsilon_0} \frac{qQ}{\zeta^2} \hat{\zeta} F = 4 π ϵ 0 1 ζ 2 qQ ζ ^ where
\vec{\zeta} = \vec{r} - \vec{r'}, \hat{\zeta} = \frac{\vec{\zeta}}{\zeta} ζ ⃗ = r ⃗ − r ′ ⃗ , ζ ^ = ζ ⃗ ζ \vec{\zeta} = \vec{r} - \vec{r'}, \hat{\zeta} = \frac{\vec{\zeta}}{\zeta} ζ = r − r ′ , ζ ^ = ζ ζ . Griffiths uses script r for this which makes more sense but I can't :( The permittivity of free space
\epsilon_0 = 8.85\text{e-12 C2/(N m2)} ϵ 0 = 8.85 e-12 C2/(N m2) \epsilon_0 = 8.85\text{e-12 C2/(N m2)} ϵ 0 = 8.85 e-12 C2/(N m2) For multiple charges,
\vec{E}(\vec{r}) = \frac{1}{4\pi \epsilon_0} \sum \frac{q}{\zeta^2} \hat{\zeta}
E ⃗ ( r ⃗ ) = 1 4 π ϵ 0 ∑ q ζ 2 ζ ^ \vec{E}(\vec{r}) = \frac{1}{4\pi \epsilon_0} \sum \frac{q}{\zeta^2} \hat{\zeta} E ( r ) = 4 π ϵ 0 1 ∑ ζ 2 q ζ ^ And for a continuous distribution,
\vec{E}(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int \frac{1}{\zeta^2} \hat{\zeta} dq
E ⃗ ( r ⃗ ) = 1 4 π ϵ 0 ∫ 1 ζ 2 ζ ^ d q \vec{E}(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int \frac{1}{\zeta^2} \hat{\zeta} dq E ( r ) = 4 π ϵ 0 1 ∫ ζ 2 1 ζ ^ d q where for a line distribution
dq = \lambda dl' d q = λ d l ′ dq = \lambda dl' d q = λ d l ′ , for a surface dq = \sigma da' d q = σ d a ′ dq = \sigma da' d q = σ d a ′ and for a volume dq = \rho d\tau' d q = ρ d τ ′ dq = \rho d\tau' d q = ρ d τ ′ . Here's a line distribution example from Griffiths:
Flux and Gauss' law Instead of calculating line, volume or surface integrals to get E fields for charge distributions, there are tricks using divergence and curl that we can use.
The flux of an E field through a surface S is:
\phi_E = \int_S \vec{E} \cdot d\vec{a}
ϕ E = ∫ S E ⃗ ⋅ d a ⃗ \phi_E = \int_S \vec{E} \cdot d\vec{a} ϕ E = ∫ S E ⋅ d a For a point charge at the origin,
\phi_E = \int_S \vec{E} \cdot d\vec{a} = \int \frac{1}{4 \pi \epsilon_0} (\frac{q}{r^2} \hat{r}) \cdot r^2 \sin \theta d \theta d \phi \hat{r} = \frac{q}{\epsilon_0}
ϕ E = ∫ S E ⃗ ⋅ d a ⃗ = ∫ 1 4 π ϵ 0 ( q r 2 r ^ ) ⋅ r 2 sin θ d θ d ϕ r ^ = q ϵ 0 \phi_E = \int_S \vec{E} \cdot d\vec{a} = \int \frac{1}{4 \pi \epsilon_0} (\frac{q}{r^2} \hat{r}) \cdot r^2 \sin \theta d \theta d \phi \hat{r} = \frac{q}{\epsilon_0} ϕ E = ∫ S E ⋅ d a = ∫ 4 π ϵ 0 1 ( r 2 q r ^ ) ⋅ r 2 sin θ d θ d ϕ r ^ = ϵ 0 q For multiple charges,
\vec{E} = \sum \vec{E_i} \implies \phi_E = \frac{Q_{enc}}{\epsilon_0}
E ⃗ = ∑ E i ⃗ ⟹ ϕ E = Q e n c ϵ 0 \vec{E} = \sum \vec{E_i} \implies \phi_E = \frac{Q_{enc}}{\epsilon_0} E = ∑ E i ⟹ ϕ E = ϵ 0 Q e n c This is Gauss' law. We can use the divergence theorem to get the differential form of Gauss' law:
\oint_S \vec{E} \cdot d\vec{a} = \int_V (\nabla \cdot \vec{E}) d\tau \\ \: \\
Q_{enc} = \int_V \rho d\tau \implies \oint_S \vec{E} \cdot d\vec{a} = \int_V \frac{\rho}{\epsilon_0} d\tau \\ \: \\
\implies \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}
∮ S E ⃗ ⋅ d a ⃗ = ∫ V ( ∇ ⋅ E ⃗ ) d τ Q e n c = ∫ V ρ d τ ⟹ ∮ S E ⃗ ⋅ d a ⃗ = ∫ V ρ ϵ 0 d τ ⟹ ∇ ⋅ E ⃗ = ρ ϵ 0 \oint_S \vec{E} \cdot d\vec{a} = \int_V (\nabla \cdot \vec{E}) d\tau \\ \: \\
Q_{enc} = \int_V \rho d\tau \implies \oint_S \vec{E} \cdot d\vec{a} = \int_V \frac{\rho}{\epsilon_0} d\tau \\ \: \\
\implies \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} ∮ S E ⋅ d a = ∫ V ( ∇ ⋅ E ) d τ Q e n c = ∫ V ρ d τ ⟹ ∮ S E ⋅ d a = ∫ V ϵ 0 ρ d τ ⟹ ∇ ⋅ E = ϵ 0 ρ Here's an application to find the E field in a coaxial cable:
There are a few more pages to this solution, see it here .
Curl of electric field and potential formation The curl of an electric field is always 0.
\nabla \times \vec{E} = 0 ∇ × E ⃗ = 0 \nabla \times \vec{E} = 0 ∇ × E = 0 . Stokes' theorem tells us that, thus, the line integral over any closed loop of an electric field is also 0. atm idk why curl is 0
This also means that the line integral from one point to another of an electric field is the same no matter the path -- the integral is path independent. (otherwise you could take one path from point A to B and another path back and end up with closed loop != 0)
Thus we can define electric potential:
V(\vec{r}) = \int_{\vec{o}}^{\vec{r}} \vec{E} d\vec{l}
V ( r ⃗ ) = ∫ o ⃗ r ⃗ E ⃗ d l ⃗ V(\vec{r}) = \int_{\vec{o}}^{\vec{r}} \vec{E} d\vec{l} V ( r ) = ∫ o r E d l where
is some standard reference point. Conventionally, is infinitely far away from what's being measured. In differential form, \vec{E} = - \nabla V
E ⃗ = − ∇ V \vec{E} = - \nabla V E = − ∇ V
Samson Zhang
