Total Force on Point Charge from 2 Point Charges

Theorem

Let $q_1$, $q_2$ and $q_3$ be point charges.

Let $\mathbf F_{ij}$ denote the force exerted on $q_j$ by $q_i$.

Let $\mathbf F_i$ denote the force exerted on $q_i$ by the combined action of the other two point charges.

Then the force $\mathbf F_1$ exerted on $q_1$ by the combined action of $q_2$ and $q_3$ is given by:

\(\ds \mathbf F_1\)

\(=\)

\(\ds \mathbf F_{21} + \mathbf F_{31}\)

\(\ds \)

\(=\)

\(\ds \dfrac {q_2 q_1} {4 \pi \varepsilon_0 {r_{2 1} }^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 {r_{3 1} }^3} \mathbf r_{3 1}\)

where:

$\mathbf F_{21} + \mathbf F_{31}$ denotes the vector sum of $\mathbf F_{21}$ and $\mathbf F_{31}$

$\mathbf r_{ij}$ 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

By definition, the force $\mathbf F_{21}$ and $\mathbf F_{31}$ are vector quantities.

Hence their resultant can be found by using the Parallelogram Law.

The result follows from Coulomb's Law of Electrostatics.

$\blacksquare$

Also presented as

This can also be presented as:

\(\ds \mathbf F_1\)

\(=\)

\(\ds \mathbf F_{21} + \mathbf F_{31}\)

\(\ds \)

\(=\)

\(\ds \dfrac {q_1} {4 \pi \varepsilon_0} \paren {\dfrac {q_2} { {r_{2 1} }^3} \mathbf r_{2 1} + \dfrac {q_3} { {r_{3 1} }^3} \mathbf r_{3 1} }\)

Also see

• Total Force on Point Charge from Multiple Point Charges

Sources

• 1990: I.S. Grant and W.R. Phillips: Electromagnetism (2nd ed.) ... (previous) ... (next): Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge