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--- 184_notes:examples:week2_electric_field_negative_point [2018/01/22 01:16] – [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:pc_efield|Return to Electric Field]]
+==== Example: Electric Field from a Negative Point Charge =====
 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===
-{{ 184_notes:2_potential_negative_point.png?150 |Negative Point Charge -Q, and Point P}}
+[{{ 184_notes:2_potential_negative_point.png?150 |Negative Point Charge -Q, and Point P}}]
+<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.
+</WRAP>
 ===Goal===
@@ Line 22: / Line 29: @@
 $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$
-We can plug in our charge ($-Q$) and the magnitude of the separation vector ($R$) to get:
+We can plug in our charge ($-Q$) and the magnitude of the separation vector (magnitude $R$) 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 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:2_electric_field_negative_point_s_direction.png?300 |s-direction drawn in}}
+[{{ 184_notes:2_electric_field_negative_point_s_direction.png?300 |s-direction drawn in}}]
-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:
+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:
 $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{R^2}\hat{s}$$
-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:
+Since the charge is negative, this means that the electric field points in the //opposite// direction of the separation vector. To make this more explicit, we could put the negative sign right next to our unit-vector:
 $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(-\hat{s})$$
 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.
+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:2_electric_field_negative_point_solution.png?150 |Charge Distribution Induced From Two Sides, Solution}}
+[{{ 184_notes:2_electric_field_negative_point_solution.png?150 |Charge Distribution Induced From Two Sides, Solution}}]