Electric Field Strength from Assemblage of Point Charges

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Theorem

Let $q_1, q_2, \ldots, q_n$ be point charges.

Let $\mathbf r_1, \mathbf r_2, \ldots, \mathbf r_n$ be the position vectors of $q_1, q_2, \ldots, q_n$ respectively.


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


Then:

$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \sum_i \dfrac {q_i} {\size {\mathbf r - \mathbf r_i}^3} \paren {\mathbf r - \mathbf r_i}$

where $\varepsilon_0$ denotes the vacuum permittivity.


Proof

Let $q$ be a test charge in the vicinity of $q_1, q_2, \ldots, q_n$ at $\map P {\mathbf r}$.

For all $i$ in $\set {1, 2, \ldots, n}$, let $\mathbf F_i$ denote the force exerted on $q$ by $q_i$.


Let $\mathbf F$ be the force exerted on $q$ by the combined action of $q_1, q_2, \ldots, q_n$.


We have:

\(\ds \mathbf F\) \(=\) \(\ds \dfrac 1 {4 \pi \varepsilon_0} \sum_i \dfrac {q q_i} {\size {\mathbf r - \mathbf r_i}^3} \paren {\mathbf r - \mathbf r_i}\) Total Force on Point Charge from Multiple Point Charges
\(\ds \leadsto \ \ \) \(\ds \dfrac {\mathbf F} q\) \(=\) \(\ds \dfrac 1 {4 \pi \varepsilon_0} \sum_i \dfrac {q_j} {\size {\mathbf r - \mathbf r_i}^3} \paren {\mathbf r - \mathbf r_i}\) dividing both sides by $q$ which is constant
\(\ds \leadsto \ \ \) \(\ds \map {\mathbf E} {\mathbf r}\) \(=\) \(\ds \dfrac 1 {4 \pi \varepsilon_0} \sum_i \dfrac {q_j} {\size {\mathbf r - \mathbf r_i}^3} \paren {\mathbf r - \mathbf r_i}\) Definition of Electric Field Strength

Hence the result.

$\blacksquare$


Sources

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