Point Charge: Difference between revisions
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'''Step 3:''' Multiply the magnitude by <math>\hat{r}</math> to find the Electric Field | '''Step 3:''' Multiply the magnitude by <math>\hat{r}</math> to find the Electric Field | ||
E= <math> \frac{1}{4 \pi \epsilon_0 } \frac{1.6 * 10^{-19}}{10}*<\frac{1}{\sqrt{10}},\frac{-3}{\sqrt{10}},\frac{0}{\sqrt{10}}> | E= <math> \frac{1}{4 \pi \epsilon_0 } \frac{1.6 * 10^{-19}}{10}*<\frac{1}{\sqrt{10}},\frac{-3}{\sqrt{10}},\frac{0}{\sqrt{10}}>=<4.554*10^{-11},-1.366^{-10},0> </math> | ||
===Middling=== | ===Middling=== | ||
Revision as of 19:29, 28 October 2015
This page is all about the Electric Field due to a Point Charge.
Electric Field
A Work In Progress by Brandon Weiner: bweiner6 (talk)
A Mathematical Model of Electric Field due to Point Charge
The Electric Field of a Point Charge can be found by the formula:
[math]\displaystyle{ \vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r,\text{where } \frac{1}{4 \pi \epsilon_0 } \text{is approximately } 9*10^{9} \text{, q is the charge of the particle, r is the magnitude of the distance between the point charge and the observation point, and } }[/math] [math]\displaystyle{ \hat r \text { is the direction of the distance from the point charge to the observation point.} }[/math]
A Computational Model
Here is a link to some code which shows the Electric Field due to an Proton at different points.
<html> <iframe src="https://trinket.io/embed/glowscript/cf036f65f7?start=result" width="100%" height="600" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe> </html>
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
Problem 1: There is a proton at <1,2,3>. Calculate the electric field at <2,-1,3>.
Step 1: Find [math]\displaystyle{ \hat r }[/math]
Find [math]\displaystyle{ \vec r_{obs} - \vec r_{proton} (\lt 2,-1,3\gt - \lt 1,2,3\gt = \lt 1,-3,0\gt ) }[/math]
Calculate the magnitude of r. ([math]\displaystyle{ \sqrt{1^2+(-3)^2+0^2}=\sqrt{10} }[/math]
From r, find the unit vector [math]\displaystyle{ \hat{r}. }[/math] [math]\displaystyle{ \lt \frac{1}{\sqrt{10}},\frac{-3}{\sqrt{10}},\frac{0}{\sqrt{10}}\gt }[/math]
Step 2: Find the magnitude of the Electric Field
[math]\displaystyle{ E= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{1.6 * 10^{-19}}{10} }[/math]
Step 3: Multiply the magnitude by [math]\displaystyle{ \hat{r} }[/math] to find the Electric Field
E= [math]\displaystyle{ \frac{1}{4 \pi \epsilon_0 } \frac{1.6 * 10^{-19}}{10}*\lt \frac{1}{\sqrt{10}},\frac{-3}{\sqrt{10}},\frac{0}{\sqrt{10}}\gt =\lt 4.554*10^{-11},-1.366^{-10},0\gt }[/math]
Middling
Difficult
Connectedness
- How is this topic connected to something that you are interested in?
I am very interested in the idea of forces and how objects interact with each other. After you calculate the Electric Filed you can easily find the Electric Force on particle exerts on another.
- How is it connected to your major?
I am a CompE major and so Electric Fields have to do with my major because when you integrate them with respect to dL, and swap the sign, you get potential difference(voltage), which is very important in circuits. As ECE majors take circuits classes, this topic is relevant to me.
- Is there an interesting industrial application?
An interesting application is ....................
History
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See also
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Further reading
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External links
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References
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