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184_notes:examples:week2_electric_field_negative_point [2018/01/22 01:25] – [Solution] tallpaul | 184_notes:examples:week2_electric_field_negative_point [2021/05/19 15:11] (current) – schram45 | ||
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- | ==== 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 $P$. | ||
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===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=== | ||
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$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{(-Q)}{R^2}\hat{r}$$ | $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{(-Q)}{R^2}\hat{r}$$ | ||
- | This leaves us to find the direction of $\hat{r}$. The first thing to do would be to draw in the 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 |
- | {{ 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: | ||
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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: |