Producing a Radiative Electric Field - Revision history

Ck: /* Middling */

2015-12-08T04:41:31Z

Middling

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	[[File:Acceleration at angle.png]]		[[File:Acceleration at angle.png]]

	If θ is 60 degrees and <math?\|a\|</math> is 8.1 * 10^14 m/s^2, what is <math>\|a_\perp\|</math>?		If the angle represented by θ is 60 degrees and <math>\|a\|</math> is 8.1 * 10^14 m/s^2, what is <math>\|a_\perp\|</math>?

	====Procedure====		====Procedure====

Ck: /* Middling */

2015-12-08T04:39:51Z

Middling

← Older revision		Revision as of 00:39, 8 December 2015
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	[[File:Acceleration at angle.png]]		[[File:Acceleration at angle.png]]

	If θ is 60 degrees and <math?\|a\|</math> is 8.1 * 10^14 m/s^2, what is <math?\|a_\perp\|</math>?		If θ is 60 degrees and <math?\|a\|</math> is 8.1 * 10^14 m/s^2, what is <math>\|a_\perp\|</math>?

	====Procedure====		====Procedure====

Ck: /* Procedure */

2015-12-08T04:39:07Z

Procedure

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	====Procedure====		====Procedure====

	Because the angle represented by θ is opposite the <math?a_\perp</math> vector, we simply multiply <math?\|a\|</math> by sin(θ) to compute our answer:		Because the angle represented by θ is opposite the <math>a_\perp</math> vector, we simply multiply <math>\|a\|</math> by sin(θ) to compute our answer:

	<math> 8.1 * 10^{14} m/s^2 * sin(60) = 7.05 * 10^{14} m/s^2</math>		<math> 8.1 * 10^{14} m/s^2 * sin(60) = 7.05 * 10^{14} m/s^2</math>

Ck: /* Procedure */

2015-12-08T04:38:39Z

Procedure

← Older revision		Revision as of 00:38, 8 December 2015
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	Because the angle represented by θ is opposite the <math?a_\perp</math> vector, we simply multiply <math?\|a\|</math> by sin(θ) to compute our answer:		Because the angle represented by θ is opposite the <math?a_\perp</math> vector, we simply multiply <math?\|a\|</math> by sin(θ) to compute our answer:

	<math> 8.1 * 10^{14} m/s^2 * sin(60) = 7.05 * 10^{14}</math>		<math> 8.1 * 10^{14} m/s^2 * sin(60) = 7.05 * 10^{14} m/s^2</math>

	===Difficult===		===Difficult===

Ck at 04:38, 8 December 2015

2015-12-08T04:38:16Z

← Older revision		Revision as of 00:38, 8 December 2015
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	===Middling===		===Middling===


			[[File:Acceleration at angle.png]]

			If θ is 60 degrees and <math?\|a\|</math> is 8.1 * 10^14 m/s^2, what is <math?\|a_\perp\|</math>?

			====Procedure====

			Because the angle represented by θ is opposite the <math?a_\perp</math> vector, we simply multiply <math?\|a\|</math> by sin(θ) to compute our answer:

			<math> 8.1 * 10^{14} m/s^2 * sin(60) = 7.05 * 10^{14}</math>

	===Difficult===		===Difficult===

Ck: /* Examples */

2015-12-08T04:32:08Z

Examples

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	===Middling===		===Middling===


			===Difficult===

	[[File:Radiative middling example.png]]		[[File:Radiative middling example.png]]
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	As for the directions, because <math>-qa_{\perp}</math> is in the -z direction, both fields also point in the -z direction.		As for the directions, because <math>-qa_{\perp}</math> is in the -z direction, both fields also point in the -z direction.

	~~===Difficult===~~

	==Connectedness==		==Connectedness==

Ck: /* Middling */

2015-12-08T03:40:02Z

Middling

← Older revision		Revision as of 23:40, 7 December 2015
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	The radiative electric field at this location is known as the traverse electric field, which always has a direction opposite the direction of <math>\vec{a}_\perp</math>. Thus, <math>\vec{E}_{radiative}</math> at the observation location points in the +y direction.		The radiative electric field at this location is known as the traverse electric field, which always has a direction opposite the direction of <math>\vec{a}_\perp</math>. Thus, <math>\vec{E}_{radiative}</math> at the observation location points in the +y direction.

	==Middling==		===Middling===

	[[File:Radiative middling example.png]]		[[File:Radiative middling example.png]]

Ck: /* Middling */

2015-12-08T03:39:47Z

Middling

← Older revision		Revision as of 23:39, 7 December 2015
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	The radiative electric field at this location is known as the traverse electric field, which always has a direction opposite the direction of <math>\vec{a}_\perp</math>. Thus, <math>\vec{E}_{radiative}</math> at the observation location points in the +y direction.		The radiative electric field at this location is known as the traverse electric field, which always has a direction opposite the direction of <math>\vec{a}_\perp</math>. Thus, <math>\vec{E}_{radiative}</math> at the observation location points in the +y direction.

	===Middling===		==Middling==

	[[File:Radiative middling example.png]]		[[File:Radiative middling example.png]]
	An electron experiences a kick with initial acceleration <math>a=3.6 * 10^~~32</math>~~ in the -z direction at a point (3, 4, -4) away from the origin. Calculate the direction and magnitude of the initial radiative field at (2, 3, 0) ~~and the electric field at (1, 1, 0). Then draw these arrows on the graph~~.		An electron experiences a kick with initial acceleration <math>a_\perp</math>=3.6 * 10^17 in the -z direction at a point (3, 4, -4) away from the origin. Calculate the direction and magnitude of the initial radiative field at the origin and at (2, 3, 0).

	====Procedure====		====Procedure====
	<math>E_{rad}=\frac{1}{4\pi\epsilon_0}\frac{-q\vec{a}}{c^2r}</math>		For <math>E_{rad}</math> at the origin,

			<math>E_{rad}=\frac{1}{4\pi\epsilon_0}\frac{-q\vec{a}_\perp}{c^2r}</math>

			<math>r=\sqrt{3^2 + 4^2 + 4^2} \approx 6.40</math>

			<math>E_{rad}=9 * 10^9\frac{e(3.6 * 10^{17})}{c^2 * 6.40} = \frac{8.1 * 10^{7}}{c^2} \approx 9.0 * 10^{-10}</math>

			Thus the magnitude of the initial radiative field at the origin is 9.0 * 10^-10 V/m.


			For <math>E_{rad}</math> at (2,3,0),

			<math>E_{rad}=\frac{1}{4\pi\epsilon_0}\frac{-q\vec{a}_\perp}{c^2r}</math>

			<math>r=\sqrt{1^2 + 1^2 + 4^2} \approx 4.24</math>

			<math>E_{rad}=9 * 10^9\frac{e(3.6 * 10^{17})}{c^2 * 4.24} = \frac{1.2 * 10^{8}}{c^2} \approx 1.36 * 10^{-9}</math>

			Thus the magnitude of the initial radiative field at (2,3,0) is 1.36 * 10^-9 V/m.


			As for the directions, because <math>-qa_{\perp}</math> is in the -z direction, both fields also point in the -z direction.

	===Difficult===		===Difficult===

Ck: /* Middling */

2015-12-08T03:05:53Z

Middling

← Older revision		Revision as of 23:05, 7 December 2015
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	[[File:Radiative middling example.png]]		[[File:Radiative middling example.png]]
	An electron experiences a kick in the -z direction at a 3 ~~meters to the right of and~~ 4 ~~meters above~~ the origin. ~~Draw an arrow representing~~ the direction and magnitude of the ~~electric~~ field at (2, 3, 0) and ~~another representing the direction and magnitude of~~ the electric field at (1, 1, 0).		An electron experiences a kick with initial acceleration <math>a=3.6 * 10^32</math> in the -z direction at a point (3, 4, -4) away from the origin. Calculate the direction and magnitude of the initial radiative field at (2, 3, 0) and the electric field at (1, 1, 0). Then draw these arrows on the graph.

			====Procedure====
			<math>E_{rad}=\frac{1}{4\pi\epsilon_0}\frac{-q\vec{a}}{c^2r}</math>

	===Difficult===		===Difficult===

Ck at 01:28, 8 December 2015

2015-12-08T01:28:08Z

← Older revision		Revision as of 21:28, 7 December 2015
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	===Middling===		===Middling===

			[[File:Radiative middling example.png]]
	An electron experiences a kick in the -z direction at a 3 meters to the right of and 4 meters above the origin. Draw an arrow representing the direction and magnitude of the electric field at (2, 3, 0) and another representing the direction and magnitude of the electric field at (1, 1, 0).		An electron experiences a kick in the -z direction at a 3 meters to the right of and 4 meters above the origin. Draw an arrow representing the direction and magnitude of the electric field at (2, 3, 0) and another representing the direction and magnitude of the electric field at (1, 1, 0).