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Electostatic + Electrodynamics Use varepsilon

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Indigo5684
2025-09-30 13:19:33 -05:00
parent 3a9785a406
commit 6e16e4af8f
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32
docs/physics/electrostatics/2-coulomb.md
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@@ -8,7 +8,7 @@ Thus, according to Coulomb's Law,
$$
\begin{align}
F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
\end{align}
$$
@@ -17,7 +17,7 @@ Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\v
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
\end{align}
$$
@@ -26,7 +26,7 @@ Now, let $Q$ be at point $\vb{r'}$. Then, the unit vector becomes $\frac{\vb{r}-
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
\end{align}
$$
@@ -35,7 +35,7 @@ With multiple charges, we can apply the *superposition principal* to see the fol
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
\end{align}
$$
@@ -44,7 +44,7 @@ We can convert this to an integral as $N$ goes to infinity.
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V'
\end{align}
$$
@@ -61,7 +61,7 @@ Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnet
$$
\begin{align}
\div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
\div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
\div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')}
\end{align}
$$
@@ -70,9 +70,9 @@ As the divergence operator does not operate on $\vb{r'}$, we see that
$$
\begin{align}
\div{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \epsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_e(\vb{r})}{\epsilon_0} \\
\div{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_e(\vb{r})}{\varepsilon_0} \\
\div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_m(\vb{r})}{\mu_0}
@@ -83,7 +83,7 @@ The curl of an electrostatic or magnetostatic is relatively simple.
$$
\begin{align}
\curl{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
\curl{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
\curl{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
\end{align}
$$
@@ -100,30 +100,30 @@ We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Add
## Section 2.4 - Electric and Magnetic Flux Densities
The electric and magnetic flux density vectors are given by $\epsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
The electric and magnetic flux density vectors are given by $\varepsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
$$
\Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
\Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
$$
Thus, applying divergence theorem,
$$
Q_e = \Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{E}} \dd V
Q_e = \Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{E}} \dd V
$$
$$
Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{H}} \dd V
Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{H}} \dd V
$$
Since $Q_e = \int_V \rho_e \dd V$ and $Q_m = \int_V \rho_m \dd V$, we see that
$$
\begin{align}
\int_V \rho_e \dd V &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\
\rho_e &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\
\int_V \rho_e \dd V &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
\rho_e &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
\int_V \rho_m \dd V &= \mu_0 \int_V \div{\vb{H}} \dd V \\
\rho_m &= \mu_0 \int_V \div{\vb{H}} \dd V \\
\end{align}

46
docs/physics/electrostatics/3-electro-magnetic-potentials.md
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@@ -31,7 +31,7 @@ $$
Strictly speaking, this is a potential difference. To find the absolute potential, assume a point charge $Q$ at the origin, and a charge $q$. We take the work as $q$ moves from $\vb{r'} = \vb{\infty}$ to $\vb{r'} = \vb{r}$. Thus,
$$
W = -q_e \frac{Q_e}{4 \pi \epsilon_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_e \frac{Q_e}{4 \pi \epsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \epsilon_0} \frac{1}{r}
W = -q_e \frac{Q_e}{4 \pi \varepsilon_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_e \frac{Q_e}{4 \pi \varepsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \varepsilon_0} \frac{1}{r}
$$
$$
@@ -41,25 +41,25 @@ $$
Letting the potential as $\vb{r} \rightarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following:
$$
V_e(\vb{r}) = \frac{Q_e}{4 \pi \epsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r}
V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r}
$$
Now, if we let the stationary charge $Q$ be located at $\vb{r'}$, we see that
$$
V_e(\vb{r}) = \frac{Q_e}{4 \pi \epsilon_0 \abs{\vb{r} - \vb{r'}}} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 \abs{\vb{r} - \vb{r'}}}
V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 \abs{\vb{r} - \vb{r'}}} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 \abs{\vb{r} - \vb{r'}}}
$$
If we allow multiple charges, this becomes
$$
V_e(\vb{r}) = \frac{1}{4\pi \epsilon_0} \sum_{i=1}^N \frac{Q_ei}{\abs{\vb{r}-\vb{r_i}}}
V_e(\vb{r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^N \frac{Q_ei}{\abs{\vb{r}-\vb{r_i}}}
$$
Taking this to its natural limit,
$$
V_e(\vb{r}) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
V_e(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
$$
$$
@@ -71,25 +71,25 @@ $$
Given two point charges $Q_{e1}, Q_{e2}$ we know the work to bring them together is
$$
W_2 = W_{21} = \frac{1}{4 \pi \epsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}}
W_2 = W_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}}
$$
Superposition applies here. The energy to create $N$ charges is
$$
W_n = \frac{1}{2} \frac{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
W_n = \frac{1}{2} \frac{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
$$
For the sake of symmetry, sum overall charges and divide by 2.
$$
W_n = \frac{1}{2} \frac{1}{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
W_n = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
$$
Rearranging, we see the following:
$$
W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \epsilon_0} \frac{Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \varepsilon_0} \frac{Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
= \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\vb{r_i})
$$
@@ -103,7 +103,7 @@ $$
We can also express this as
$$
W_e = \frac{1}{2} \frac{1}{4 \pi \epsilon_0} \int_V \int_{V'} \frac{\rho_e(\vb{r})\rho_e(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V}
W_e = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \int_V \int_{V'} \frac{\rho_e(\vb{r})\rho_e(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V}
$$
$$
@@ -115,13 +115,13 @@ Note the $\frac{1}{2}$ is the same anti-double-counting factor introduced previo
We can now write an expression for energy of a charge density in terms of the field that it produces.
$$
W = \frac{\epsilon_0}{2} \int_V (\div{\vb{E}(\vb{r})}) V(\vb{r}) \dd V
W = \frac{\varepsilon_0}{2} \int_V (\div{\vb{E}(\vb{r})}) V(\vb{r}) \dd V
$$
Simplifying, we see that
$$
W_e = \frac{\epsilon_0}{2} \int_{V} E^2(\vb{r}) \dd V ; \quad
W_e = \frac{\varepsilon_0}{2} \int_{V} E^2(\vb{r}) \dd V ; \quad
W_m = \frac{\mu_0}{2} \int_{V} H^2(\vb{r}) \dd V
$$
@@ -132,7 +132,7 @@ We know that $\vb{E}(\vb{r}) = -\div{V_e(\vb{r})}$ and $\vb{H}(\vb{r}) = -\div{V
Combined this, as well as the first of the Maxwell equations, we see that
$$
\div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\epsilon_0}
\div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\varepsilon_0}
$$
$$
@@ -150,13 +150,13 @@ $$
Now, we can construct a potential function in terms of said green function that satisfies the Laplace equation.
$$
V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\epsilon_0} \dd{V'}
V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'}
$$
This is the specific solution. Let $\psi(\vb{r})$ be a solution to the homogenous equation. We can state the following:
$$
V_e(\vb{r}) = \psi(\vb{r}) - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\epsilon_0} \dd{V'}
V_e(\vb{r}) = \psi(\vb{r}) - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'}
$$
We will consider the potential of a point charge. THat is, the limit of potential is zero as distance approaches infinity.
@@ -164,20 +164,20 @@ We will consider the potential of a point charge. THat is, the limit of potentia
Recall the potential of a point charge:
$$
V_e(\vb{r}) = \frac{Q_e}{\epsilon_0} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}}
V_e(\vb{r}) = \frac{Q_e}{\varepsilon_0} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}}
$$
We know that $- \laplacian{V(\vb{r})} = \div{\vb{E}(\vb{r})}$. Thus, recall the electric field of a point charge.
$$
\vb{E}(\vb{r}) = -\grad{V(\vb{r})} = \frac{Q_e}{\epsilon_0} \grad({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) = \frac{Q_e}{\epsilon_0} \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3}
\vb{E}(\vb{r}) = -\grad{V(\vb{r})} = \frac{Q_e}{\varepsilon_0} \grad({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) = \frac{Q_e}{\varepsilon_0} \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3}
$$
Taking the divergence, we find that
$$
- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\epsilon_0} = \frac{Q_e}{\epsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}})
= \frac{Q_e}{\epsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\epsilon_0} \delta(\vb{r} - \vb{r'})
- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\varepsilon_0} = \frac{Q_e}{\varepsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}})
= \frac{Q_e}{\varepsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\varepsilon_0} \delta(\vb{r} - \vb{r'})
$$
Thus, we see that
@@ -189,7 +189,7 @@ $$
Finally,
$$
V_e(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_e}{\epsilon_0} \dd{V'}
V_e(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_e}{\varepsilon_0} \dd{V'}
$$
$$
@@ -230,9 +230,9 @@ $$
Applying this to the equation for voltage, we see that
$$
V_e(r) = \frac{1}{\epsilon_0} \int_{V'} G(\vb{r}, \vb{r'}) p_e(\vb{r'}) \dd{V'} \approx
\frac{1}{4 \pi \epsilon_0 r} \int_{V'} (1 + \frac{\vu{r} \vdot \vb{r'}}{r}) p_e(\vb{r'}) \dd{V'}
= \frac{Q_e}{4 \pi \epsilon_0 r} + \frac{\vu{r} \vdot \vb{p}}{4 \pi \epsilon_0 r^2}
V_e(r) = \frac{1}{\varepsilon_0} \int_{V'} G(\vb{r}, \vb{r'}) p_e(\vb{r'}) \dd{V'} \approx
\frac{1}{4 \pi \varepsilon_0 r} \int_{V'} (1 + \frac{\vu{r} \vdot \vb{r'}}{r}) p_e(\vb{r'}) \dd{V'}
= \frac{Q_e}{4 \pi \varepsilon_0 r} + \frac{\vu{r} \vdot \vb{p}}{4 \pi \varepsilon_0 r^2}
$$
**Definition**. The first and second terms of this equation are the *monopole* and *dipole* terms respectively.

10
docs/physics/electrostatics/4-conductors.md
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@@ -16,7 +16,7 @@ Consider any two points internal to the conductor. The voltage between said poin
The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The parallel voltage difference is zero, so the electric field must be zero.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\epsilon_0} = \frac{A\sigma_e}{\epsilon_0}$. Thus, $\sigma_e = \epsilon_0 E$.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$.
## Section 4.3 - Exercises involving conductors at fixed potentials
@@ -48,19 +48,19 @@ This field is zero outside of a material, and if non-zero inside a material, wil
**Definition**. Charges not bound are called *free*, with density denoted as $\rho_{ef}$. Combined with $\rho_{eb}$, they form the basis of the electric field. THat is,
$$
\epsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb}
\varepsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb}
$$
**Definition**. The electric flux density field $\vb{D}$ is defined as
$$
\vb{D} = \epsilon_0 \vb{E} + \vb{P}
\vb{D} = \varepsilon_0 \vb{E} + \vb{P}
$$
Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that
$$
\div \vb{D} = \div (\epsilon_0 \vb{E} + \vb{P}) = \div \epsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
\div \vb{D} = \div (\varepsilon_0 \vb{E} + \vb{P}) = \div \varepsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
$$
## Section 4.5 - Induced Electric Charges, their Potentials and Fields
@@ -75,7 +75,7 @@ $$
C := \frac{Q}{V}
$$
This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \epsilon_0 R$. For a parallel plate capacitor, this reduces to $C = \frac{epsilon_0 A}{d}$.
This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \varepsilon_0 R$. For a parallel plate capacitor, this reduces to $C = \frac{epsilon_0 A}{d}$.
## Section 4.7 - Forces on Charged Conductors in Electric Fields