Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 184_notes:examples:week2_electric_field_negative_point [2018/01/22 01:26] – [Solution] tallpaul | 184_notes:examples:week2_electric_field_negative_point [2021/05/19 15:11] (current) – schram45 | ||
|---|---|---|---|
| Line 1: | Line 1: | ||
| - | ==== Electric Field from a Negative Point Charge ===== | + | [[184_notes: |
| + | ==== Example: | ||
| Suppose we have a negative 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 $P$. | Suppose we have a negative 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 $P$. | ||
| Line 8: | Line 9: | ||
| ===Representations=== | ===Representations=== | ||
| - | {{ 184_notes: | + | [{{ 184_notes: |
| + | |||
| + | <WRAP TIP> | ||
| + | ===Assumptions=== | ||
| + | * Constant charge: Makes charge in electric field equation not dependent on time or space as no information is given in problem suggesting so. | ||
| + | * Charge is not moving: This makes our separation vector fixed in time as a moving charge would have a changing separation vector with time. | ||
| + | </ | ||
| ===Goal=== | ===Goal=== | ||
| Line 27: | Line 34: | ||
| This leaves us to find the unit vector $\hat{r}$. The first thing to do would be to draw in the separation 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 (our observation location). We need to define a set of coordinate axes. 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 separation vector (since it points in some diagonal direction). | This leaves us to find the unit vector $\hat{r}$. The first thing to do would be to draw in the separation 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 (our observation location). We need to define a set of coordinate axes. 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 separation vector (since it points in some diagonal direction). | ||
| - | {{ 184_notes: | + | [{{ 184_notes: |
| Instead, we'll pick a coordinate direction that falls along the same axis as the separation vector, $\vec{r}_{-Q\rightarrow P}$. 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: | Instead, we'll pick a coordinate direction that falls along the same axis as the separation vector, $\vec{r}_{-Q\rightarrow P}$. 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: | ||
| Line 36: | Line 43: | ||
| This gives the magnitude of the electric field as $$\left|\vec{E}\right| = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}$$ | 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 $-\hat{s}$. 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 indicating the direction of the electric field is shown below. | with the direction is given by $-\hat{s}$. 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 indicating the direction of the electric field is shown below. | ||
| - | {{ 184_notes: | + | [{{ 184_notes: |