PX157 - B2a - electric field from a continuous distribution

method

divide the distribution into small elements at ${\vec{r}}^{'}$
find $d \vec{E}$ at position, $\vec{r}$ , for each element
vector sum all contributions: $\vec{E} (\vec{r}) = \int d \vec{E}$

linear charge density, $λ$

λ = \frac{d q}{d l}

\begin{aligned} d \vec{E} & = \frac{d q}{4 π ϵ_{0}} \frac{\vec{r} - {\vec{r}}^{'}}{| \vec{r} - {\vec{r}}^{'} |^{3}} \\ = \frac{λ d l}{4 π ϵ_{0}} \frac{\vec{r} - {\vec{r}}^{'}}{| \vec{r} - {\vec{r}}^{'} |^{3}} \\ \vec{E} & = \frac{1}{4 π ϵ_{0}} \int λ . d l \frac{\vec{r} - {\vec{r}}^{'}}{| \vec{r} - {\vec{r}}^{'} |^{3}} \end{aligned}

surface charge density, $σ$

σ = \frac{d q}{d S}

\begin{aligned} d \vec{E} & = \frac{σ d S}{4 π ϵ_{0}} \frac{\vec{r} - {\vec{r}}^{'}}{| \vec{r} - {\vec{r}}^{'} |^{3}} \\ \vec{E} & = \frac{1}{4 π ϵ_{0}} \iint σ . d S \frac{\vec{r} - {\vec{r}}^{'}}{| \vec{r} - {\vec{r}}^{'} |^{3}} \end{aligned}

volume charge density, $ρ$

ρ = \frac{d q}{d τ}

\begin{aligned} d \vec{E} & = \frac{ρ d τ}{4 π ϵ_{0}} \frac{\vec{r} - {\vec{r}}^{'}}{| \vec{r} - {\vec{r}}^{'} |^{3}} \\ \vec{E} & = \frac{1}{4 π ϵ_{0}} ∭ ρ . d τ \frac{\vec{r} - {\vec{r}}^{'}}{| \vec{r} - {\vec{r}}^{'} |^{3}} \end{aligned}

eg: a uniformly distributed charged circle, centred at the origin along the $y - z$ plane, with total charge, $Q$ $d \vec{E} (x, 0, 0) = \frac{d Q}{4 π ϵ_{0}} \frac{x \hat{x} - {\vec{r}}^{'}}{| x \hat{x} - {\vec{r}}^{'} |^{3}}$ $\vec{E} (x, 0, 0) = \int \frac{d Q}{4 π ϵ_{0}} \frac{x \hat{x} - {\vec{r}}^{'}}{| x \hat{x} - {\vec{r}}^{'} |^{3}}$

\begin{aligned} E_{x} (x, 0, 0) & = (\vec{E} \cdot \hat{x}) \\ = \int \frac{x}{4 π ϵ_{0} r} d Q \\ = \frac{x}{4 π ϵ_{0} r} \int d Q \\ = \frac{x Q}{4 π ϵ_{0} r} \\ = \frac{x Q}{4 π ϵ_{0} (x^{2} + a^{2})^{\frac{3}{2}}} \end{aligned}

also, $E_{x} (0, 0, 0) = 0$ , and $lim_{x \to \pm \infty} E_{x} = 0$