Magnitude of Electric Field caused by Point Charge

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Theorem

Let $q$ be a point charge.

Let $\map {\mathbf E} {\mathbf r}$ be the electric field strength due to $q$ at a point $P$ whose position vector is $\mathbf r$.


The magnitude of the electric field strength due to $q$ at $P$ is given by:


Then:

$\size {\map {\mathbf E} {\mathbf r} } = \dfrac {\size q} {4 \pi \epsilon_0 r^2}$

where:

$r$ is the distance $P$ is from $q$
$\varepsilon_0$ denotes the vacuum permittivity.


Proof

\(\ds \map {\mathbf E} {\mathbf r}\) \(=\) \(\ds \dfrac 1 {4 \pi \epsilon_0} \dfrac {q \paren {\mathbf r - \mathbf r_q} } {\size {\mathbf r - \mathbf r_q}^3}\) Electric Field caused by Point Charge
\(\ds \leadsto \ \ \) \(\ds \size {\map {\mathbf E} {\mathbf r} }\) \(=\) \(\ds \size {\dfrac 1 {4 \pi \epsilon_0} \dfrac {q \paren {\mathbf r - \mathbf r_q} } {\size {\mathbf r - \mathbf r_q}^3} }\)
\(\ds \) \(=\) \(\ds \dfrac {\size q} {4 \pi \epsilon_0} \size {\dfrac {\mathbf r - \mathbf r_q} {\size {\mathbf r - \mathbf r_q}^3} }\)
\(\ds \) \(=\) \(\ds \dfrac {\size q} {4 \pi \epsilon_0} \dfrac {\size {\mathbf r - \mathbf r_q} } {\size {\mathbf r - \mathbf r_q}^3}\)
\(\ds \) \(=\) \(\ds \dfrac {\size q} {4 \pi \epsilon_0} \dfrac 1 {\size {\mathbf r - \mathbf r_q}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {\size q} {4 \pi \epsilon_0 r^2}\)


The following diagram shows the points of equal magnitude of $\map {\mathbf E} {\mathbf r}$ as dotted circles:

Field-lines-positive-charge.png

$\blacksquare$


Sources