Suppose we have a negative charge with charge $-Q$. What is the magnitude of the electric field at a point $P$, which is a distance $R$ from the charge? Draw the electric field vector on a diagram to show the direction of the electric field at $P$.

• The point $P$ is a distance $R$ away from the point charge.
• The separation vector $\vec r$ points from $-Q$ to $P$, and has magnitude $R$.

• Magnitude of the electric field at $P$.
• Direction of the electric field at $P$.

• The charge with value $-Q$ is a point charge.
• The electric field at $P$ is made up entirely of contributions from the point charge (there are no other surrounding charges).

• The electric field from the point charge can be written as $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r},$$ where $q$ represents our charge ($-Q$ in this case), $r$ is our distance ($R$), and $\hat r$ is the unit vector pointing from the point charge to $P$.
• We can represent the electric field in our diagram with an arrow, since the electric field at a specific point is a vector.

The electric field at $P$ is given by: $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$

We can plug in our charge ($-Q$) and the magnitude of the separation vector ($R$) to get: $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{(-Q)}{R^2}\hat{r}$$

This leaves us to find the direction of $\hat{r}$. The first thing to do would be to draw in the vector $\vec{r}_{-Q\rightarrow P}$. This vector points from the charge -Q to Point P since P is where we want to find the electric field. Now we need to define a coordinate axis. We could pick the normal x- and y-axes, but this would make writing the $\vec{r}$ and $\hat{r}$ more difficult because there would be both x and y components to the r-vector (since it points in some diagonal direction).

Instead, let's pick a coordinate direction that falls along the same axis as the $\vec{r}_{-Q\rightarrow P}$. Since this is a coordinate direction that we're naming, let's call this the $\hat{s}$ direction. That means that $\vec{r}_{-Q\rightarrow P}$ points in the $\hat{s}$ direction, so $\hat{r}=\hat{s}$. Plugging this into our electric field equation gives: $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{R^2}\hat{s}$$ Since the charge is negative, this means that the electric field points in the opposite direction of the $\vec{r}$. To make this explicit, we could write this as: $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(-\hat{s})$$

This gives the magnitude of the electric field as $$\left|\vec{E}\right| = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}$$ with the direction is given by $-\hat{s}$, which is the opposite of the direction of $\vec{r}_{-Q \rightarrow P}$. The electric field at $P$ therefore points from $P$ to the point charge. You'll find this to be true for all negative charge - the electric field points towards negative charges. A diagram is shown below.