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--- 184_notes:examples:week2_electric_field_negative_point [2017/08/28 20:35] – [Solution] tallpaul
+++ 184_notes:examples:week2_electric_field_negative_point [2021/05/19 15:11] (current) – schram45
@@ Line 1: / Line 1: @@
-===== Example: Electric Field from a Negative Point Charge =====
+[[184_notes:pc_efield|Return to Electric Field]]
-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 $P$.
+==== Example: Electric Field from a Negative Point Charge =====
-{{ 184_notes:2_potential_negative_point.png?150 |Negative Point Charge -Q, and Point 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 $P$.
 ===Facts===
-  * The point $P$ is a distance $R$ away from the point charge.
+  * There is a negative charge $-Q$.
-  * The separation vector $\vec r$ points from $-Q$ to $P$, and has magnitude $R$.
+  * The point $P$ is a distance $R$ away from the charge, with the orientation shown below in our representation.
+  * The electric field from the point charge at a particular observation location can be written as $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r},$$ where $q$ represents the charge, $r$ is the distance, and $\hat r$ is the unit-vector pointing from the point charge source to the observation location.
-===Lacking===
+===Representations===
-  * Magnitude of the electric field at $P$.
+[{{ 184_notes:2_potential_negative_point.png?150 |Negative Point Charge -Q, and Point P}}]
-  * Direction of the electric field at $P$.
-===Approximations & Assumptions===
+<WRAP TIP>
-  * The charge with value $-Q$ is a point charge.
+===Assumptions===
-  * The electric field at $P$ is made up entirely of contributions from the point charge (there are no other surrounding charges).
+  * 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>
-===Representations===
+===Goal===
-  * 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 $P$.
+  * Find the magnitude and direction of the electric field at $P$.
-  * We can represent the electric field in our diagram with an arrow, since the electric field at a specific point is a vector.
 ====Solution====
-The electric field at $P$ is given by:
+<WRAP TIP>
+=== Approximation ===
+We approximate the charge with value $-Q$ as a point charge. This may not be the case, but our approximation does not seem unreasonable based on the limited information. In fact, this approximation is necessary, since we do not yet know how to represent electric field apart from the electric field from a point charge. We proceed with this in mind.
+</WRAP>
+The electric field at $P$ is given by the electric field from a point charge:
 $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$
-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 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).
+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}}]