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:
$\blacksquare$
Sources
- 1990: I.S. Grant and W.R. Phillips: Electromagnetism (2nd ed.) ... (previous) ... (next): Chapter $1$: Force and energy in electrostatics: $1.2$ The Electric Field