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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}}]