184_notes:pc_efield

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184_notes:pc_efield [2021/01/25 01:50] – bartonmo184_notes:pc_efield [2021/05/26 13:39] (current) – schram45
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 ==== Electric Field Vectors ==== ==== Electric Field Vectors ====
  
-[{{  184_notes:efieldvectora.png?250|Electric Field at observation point A from Q}}] +[{{ :184_notes:efieldvectora_new.png?250|Electric Field at observation point A from Q}}] 
-To understand the electric field around a point charge (or any other distribution of charge), we will often draw vectors around the charge called "electric field vectors" or just "field vectors." **The magnitude or length of these vectors represents the magnitude of the electric field, and the direction of the vector points in the same direction as the electric field** (shown with the blue $\vec{E_A}$ arrow in the figure on the right).+To understand the electric field around a point charge (or any other distribution of charge), we will often draw vectors around the charge called "electric field vectors" or just "field vectors." **The magnitude or length of these vectors represents the magnitude of the electric field, and the direction of the vector points in the same direction as the electric field** (shown with the dashed blue $\vec{E_A}$ arrow in the figure on the right).
  
-For a positive point charge Q, consider Points A-D, each a distance d (shown in red) from the charge. To draw the electric field vectors around this charge, we need to find the magnitude **//and//** direction of the electric field at each point. Starting with the electric field equation, we can find the electric field for Point A. We already know that the charge is Q, so we have:+For a positive point charge Q, consider Points A-D, each a distance d from the charge. To draw the electric field vectors around this charge, we need to find the magnitude **//and//** direction of the electric field at each point. Starting with the electric field equation, we can find the electric field for Point A. We already know that the charge is Q, so we have:
 $$\vec{E_A} = \frac{1}{4 \pi\epsilon_0}\frac{Q}{r_A^2} \hat{r_A}$$    $$\vec{E_A} = \frac{1}{4 \pi\epsilon_0}\frac{Q}{r_A^2} \hat{r_A}$$   
  
-All we need now is to find the separation vector $\vec{r_A}$, which will help us find the magnitude $r_A$ and unit vector $\hat{r_A}$. The separation vector $\vec{r_A}$ will point from the charge to Point A (drawn as the red arrow in the figure). Using the traditional cartesian coordinate system (+x points to the right, +y points up, the origin at the positive point charge), then $\vec{r_A}$ points only in the +y-direction, so $\vec{r_A} = \langle 0, d, 0 \rangle$. The magnitude of $\vec{r_A}$ is then:+All we need now is to find the separation vector $\vec{r_A}$, which will help us find the magnitude $r_A$ and unit vector $\hat{r_A}$. The separation vector $\vec{r_A}$ will point from the charge to Point A (drawn as the solid red arrow in the figure). Using the traditional cartesian coordinate system (+x points to the right, +y points up, the origin at the positive point charge), then $\vec{r_A}$ points only in the +y-direction, so $\vec{r_A} = \langle 0, d, 0 \rangle$. The magnitude of $\vec{r_A}$ is then:
 $$r_A=|\vec{r_A}|=\sqrt{r_{Ax}^2+r_{Ay}^2+r_{Az}^2}=\sqrt{0^2+d^2+0^2}$$ $$r_A=|\vec{r_A}|=\sqrt{r_{Ax}^2+r_{Ay}^2+r_{Az}^2}=\sqrt{0^2+d^2+0^2}$$
 $$r_A=d$$  $$r_A=d$$ 
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 $$\vec{E_A} = \frac{1}{4 \pi\epsilon_0}\frac{Q}{d^2} \hat{y}$$ $$\vec{E_A} = \frac{1}{4 \pi\epsilon_0}\frac{Q}{d^2} \hat{y}$$
  
-[{{  184_notes:efieldvectors.png?200|Electric Field from a point charge}}]+[{{ :184_notes:efieldvectors_new.png?250|Electric Field from a point charge}}]
 So we draw the electric field vector at Point A pointing straight up. If you follow the same steps for Points B-D, you find an important pattern from drawing this electric field vectors: **the electric field from a positive point charge points away from the charge**. If we were to look at points that were a distance of 2d away from the point charge, we would need to change the magnitude of the electric field by a factor of 4 (since it is $r^2$ in the denominator), but the directions would stay the same. So we draw the electric field vector at Point A pointing straight up. If you follow the same steps for Points B-D, you find an important pattern from drawing this electric field vectors: **the electric field from a positive point charge points away from the charge**. If we were to look at points that were a distance of 2d away from the point charge, we would need to change the magnitude of the electric field by a factor of 4 (since it is $r^2$ in the denominator), but the directions would stay the same.
  
 ==== Examples ==== ==== Examples ====
-[[184_notes:examples:Week2_electric_field_negative_point|Electric Field from a Negative Point Charge]]+  * [[184_notes:examples:Week2_electric_field_negative_point|Electric Field from a Negative Point Charge]] 
 +    * Video Example: Electric Field from a Negative Point Charge 
 +{{youtube>a64SCwLdIe0?large}}
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  • Last modified: 2021/01/25 01:50
  • by bartonmo