Direction of Electric Field caused by Point Charge

Theorem

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 direction of the electric field due to $q$ at $P$ is:

for positive $q$, directly away from $q$

for negative $q$, directly towards $q$.

$\ds \map {\mathbf E} {\mathbf r}$

$=$

$\ds \dfrac 1 {4 \pi \epsilon_0} \dfrac {q \paren {\mathbf r - \mathbf r_q} } {\size {\mathbf r - \mathbf r_q}^3}$

By definition of vector subtraction, $\mathbf r - \mathbf r_q$ is the vector from $\mathbf r_q$ to $\mathbf r$.

Hence if $q$ is positive, the direction of $\map {\mathbf E} {\mathbf r}$ is towards $\mathbf r$, which is directly away from $q$.

Similarly, if $q$ is negative, the direction of $\map {\mathbf E} {\mathbf r}$ is $-\paren {\mathbf r - \mathbf r_q}$.

That is, the direction of $\map {\mathbf E} {\mathbf r}$ is $\mathbf r_q - \mathbf r$.

By definition of vector subtraction, $\mathbf r_q - \mathbf r$ is the vector from $\mathbf r$ to $\mathbf r_q$.

This is directly towards $q$.

The following diagram shows the lines of force of $\map {\mathbf E} {\mathbf r}$ as arrows, pointing away from $+q$:

The following diagram shows the lines of force of $\map {\mathbf E} {\mathbf r}$ as arrows, pointing towards $-q$:

$\blacksquare$

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