Direction of Electric Field caused by Point Charge

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


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


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

Proof
From Electric Field caused by Point Charge:

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 $+q$:


 * Field-lines-positive-charge.png

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


 * Field-lines-negative-charge.png