Total Force on Point Charge from Multiple Point Charges

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Theorem

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

For all $i$ in $\set {1, 2, \ldots, n}$ where $i \ne j$, let $\mathbf F_{i j}$ denote the force exerted on $q_j$ by $q_i$.

For all $i$ in $\set {1, 2, \ldots, n}$, let $\mathbf F_i$ denote the force exerted on $q_i$ by the combined action of all the other point charges.


Then the electrostatic force $\mathbf F_i$ exerted on $q_i$ by the combined action of all the other point charges is given by:

\(\ds \mathbf F_i\) \(=\) \(\ds \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \mathbf F_{j i}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}\)

where:

the summation denotes the vector sum of $\mathbf F_{ji}$
$\mathbf r_{ij}$ denotes the displacement from $q_i$ to $q_j$
$r_{ij}$ denotes the distance between $q_i$ and $q_j$
$\varepsilon_0$ denotes the vacuum permittivity.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$


$\map P 2$ is the case: $\mathbf F_1 = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_1 q_2} {r_{2 1}^3} \mathbf r_{2 1}$

which is Coulomb's Law of Electrostatics.

Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 3$ is the case:

$\mathbf F_1 = \dfrac {q_2 q_1} {4 \pi \varepsilon_0 r_{2 1}^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 r_{3 1}^3} \mathbf r_{3 1}$

which is proved in Total Force on Point Charge from 2 Point Charges.

Thus $\map P 3$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le k \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$


from which it is to be shown that:

$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le {k + 1} \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$


Induction Step

This is the induction step:

\(\ds \mathbf F_i\) \(=\) \(\ds \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le k \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i} + \dfrac {q_i q_{k + 1} } {4 \pi \varepsilon_0 r_{\paren {k + 1} i}^3} \mathbf r_{\paren {k + 1} i}\) Basis for the Induction
\(\ds \) \(=\) \(\ds \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le {k + 1} \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}\) Definition of Summation

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 2}: \ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$


$\blacksquare$


Also presented as

This can also be presented as:

$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {\size {\mathbf r_j - \mathbf r_i}^3} \paren {\mathbf r_j - \mathbf r_i}$

where $\mathbf r_i$ denotes the position vector of $q_i$ with respect to the origin of the coordinate system used as a frame of reference.


Sources

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