Electric Field - Revision history

Pvishnubhotla6: /* A Computational Model */

2025-12-01T03:39:24Z

A Computational Model

← Older revision		Revision as of 23:39, 30 November 2025
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	[[File:EFieldArrows2025.png\|center]]		[[File:EFieldArrows2025.png\|center]]

	The following code can be used to visualize the electric field produced by a positive point charge. The interactive viewer may be found here: [https://trinket.io/glowscript/c7521c3fbc76 New field simulation code]. Note that this code was not used to generate the exact images on this page. However, it is functionally identical ~~to the old code~~ and produces ~~very similar images~~. For reference, the old code may be found at the bottom of the page or this link: [https://trinket.io/glowscript/fddb68480031 Old field simulation code].		The following code can be used to visualize the electric field produced by a positive point charge. The interactive viewer may be found here: [https://trinket.io/glowscript/c7521c3fbc76 New field simulation code]. Note that this code was not used to generate the exact images on this page. However, it is functionally identical and produces the same image; follow the link to find out for yourself. For reference, the old code may be found at the bottom of the page or this link: [https://trinket.io/glowscript/fddb68480031 Old field simulation code].
	[[File:NormalEField.png\|right\|250px\|thumb\|Normal view of simulated electric field (Laurence, 2020)]]		[[File:NormalEField.png\|right\|250px\|thumb\|Normal view of simulated electric field (Laurence, 2020)]]
	[[File:CenteredAndDistantEField.png\|right\|250px\|thumb\|Distant view of simulated electric field (Laurence, 2020)]]		[[File:CenteredAndDistantEField.png\|right\|250px\|thumb\|Distant view of simulated electric field (Laurence, 2020)]]

Pvishnubhotla6: /* A Computational Model */

2025-11-30T19:42:51Z

A Computational Model

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Pvishnubhotla6: /* Using Electric Potential */

2025-11-30T19:31:39Z

Using Electric Potential

← Older revision		Revision as of 15:31, 30 November 2025
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	====Using Electric Potential====		====Using Electric Potential====
			This is a little bit of additional information on Electric Potential:

	The electric field is also related to the electric potential <math>V</math> (not discussed at the start of the course): <math>\Delta V = V(\vec{r_{f}}) - V(\vec{r_{i}}) = -\int_{\vec{r_{i}}}^{\vec{r_{f}}} \vec{E}(\vec{r}) \cdot \vec{dl} = - \int_{x_{i}}^{x_{f}} E_{x}dx - \int_{y_{i}}^{y_{f}} E_{y}dy - \int_{z_{i}}^{z_{f}} E_{z}dz \Rightarrow		The electric field is also related to the electric potential <math>V</math> (not discussed at the start of the course): <math>\Delta V = V(\vec{r_{f}}) - V(\vec{r_{i}}) = -\int_{\vec{r_{i}}}^{\vec{r_{f}}} \vec{E}(\vec{r}) \cdot \vec{dl} = - \int_{x_{i}}^{x_{f}} E_{x}dx - \int_{y_{i}}^{y_{f}} E_{y}dy - \int_{z_{i}}^{z_{f}} E_{z}dz \Rightarrow
	\vec{E} = - \nabla V = <- \frac{\partial V}{\partial x}, - \frac{\partial V}{\partial y}, - \frac{\partial V}{\partial z}></math>		\vec{E} = - \nabla V = <- \frac{\partial V}{\partial x}, - \frac{\partial V}{\partial y}, - \frac{\partial V}{\partial z}></math>

Pvishnubhotla6: /* Connectedness */

2025-11-30T19:29:48Z

Connectedness

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 *'''Military and Defense:'''<br>
 :'''Electric Fields''' are commonly used in various weapons platforms. Weapons used to rely primarily on explosives; however, electric weapons use stored electrical energy to attack targets. There are two general types: directed-energy weapons (DEWs) and electromagnetic (EM) weapons. DEWs include lasers, radio frequency weapons, and more. EM weapons include rail guns, coil guns, etc. For example, rail guns use EM force to launch high velocity projectiles at a target. They work by using very high electrical currents to induce magnetic fields that accelerate a projectile to extremely high speeds (up to Mach 6).
 ==History==

Pvishnubhotla6: /* Non-Coulomb Electric Field */

2025-11-30T19:28:37Z

Non-Coulomb Electric Field

← Older revision		Revision as of 15:28, 30 November 2025
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	(The negative sign indicates direction.)		(The negative sign indicates direction.)

			Note: Notice how we worked the example by using other factors, such as the radius of the solenoid itself and the B-field the solenoid creates, to find the induced electric field. Non-Coulomb electric fields come only as a result of changing B fields, which is why magnetic flux is taken into account.

	==Connectedness==		==Connectedness==

Pvishnubhotla6: /* Non-Coulomb Electric Field */

2025-11-30T19:26:53Z

Non-Coulomb Electric Field

← Older revision		Revision as of 15:26, 30 November 2025
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	===Non-Coulomb Electric Field===		===Non-Coulomb Electric Field===
	'''Example:'''		'''Example:'''

	A long solenoid has radius <math>R = 0.10 \text{ m}</math> and contains 500 turns per meter.		A long solenoid has radius <math>R = 0.10 \text{ m}</math> and contains 500 turns per meter.
	The current through the solenoid increases according to		The current through the solenoid increases according to

Pvishnubhotla6: /* Non-Coulomb Electric Field */

2025-11-30T19:26:42Z

Non-Coulomb Electric Field

← Older revision		Revision as of 15:26, 30 November 2025
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	===Non-Coulomb Electric Field===		===Non-Coulomb Electric Field===
	Example:		'''Example:'''
	A long solenoid has radius <math>R = 0.10 \text{ m}</math> and contains 500 turns per meter.		A long solenoid has radius <math>R = 0.10 \text{ m}</math> and contains 500 turns per meter.
	The current through the solenoid increases according to		The current through the solenoid increases according to
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	----		----
	Solution		'''Solution'''

	'''Step 1: Find magnetic field inside the solenoid'''		'''Step 1: Find magnetic field inside the solenoid'''

Pvishnubhotla6: /* Non-Coulomb Electric Field */

2025-11-30T19:26:22Z

Non-Coulomb Electric Field

← Older revision		Revision as of 15:26, 30 November 2025
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	----		----
	~~'''~~Solution~~'''~~		Solution

	'''Step 1: Find magnetic field inside the solenoid'''		'''Step 1: Find magnetic field inside the solenoid'''
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	<math>B(t) = 4\pi \times 10^{-4} t \ \text{T}</math>		<math>B(t) = 4\pi \times 10^{-4} t \ \text{T}</math>

	~~----~~

	'''Step 2: Take derivative of magnetic field'''		'''Step 2: Take derivative of magnetic field'''

	<math>\frac{dB}{dt} = 4\pi \times 10^{-4} \ \text{T/s}</math>		<math>\frac{dB}{dt} = 4\pi \times 10^{-4} \ \text{T/s}</math>

	~~----~~

	'''Step 3: Apply Faraday’s Law'''		'''Step 3: Apply Faraday’s Law'''

Pvishnubhotla6: /* Non-Coulomb Electric Field */

2025-11-30T19:25:42Z

Non-Coulomb Electric Field

← Older revision		Revision as of 15:25, 30 November 2025
Line 254:		Line 254:

	===Non-Coulomb Electric Field===		===Non-Coulomb Electric Field===
	\~~textbf~~{~~Example:~~}		Example:
			A long solenoid has radius <math>R = 0.10 \text{ m}</math> and contains 500 turns per meter.
			The current through the solenoid increases according to
			<math>I(t) = 2.0t \ \text{A}</math>,
			where <math>t</math> is in seconds.

	~~A long solenoid has radius $R~~ = 0.10 \text{ m}~~$ and contains $500$ turns per meter.~~		Find the induced non-Coulomb electric field <math>\vec{E}_{nc}</math> at <math>r = 0.20 \text{ m}</math> outside the solenoid.
	~~The current through~~ the solenoid ~~increases according to~~
	\[
	~~I(t) = 2.0t \ \text{A},~~
	\]
	~~where $t$ is in seconds~~.

	~~Find the induced non~~-~~Coulomb electric field $\vec{E}_{\text{nc}}$ at $r = 0.20$ m~~		----
	~~outside the solenoid.~~		'''Solution'''

	~~\hrulefill~~		'''Step 1: Find magnetic field inside the solenoid'''

	\~~textbf{Step 1: Find magnetic field inside the solenoid}~~		<math>B(t) = \mu_0 n I(t)</math>

	\[		<math>B(t) = (4\pi \times 10^{-7})(500)(2t)</math>
	B(t) = \~~mu_0 n I~~(t)
	\]

	\[		<math>B(t) = 1000\pi \times 10^{-7} t</math>
	B(t) = (4\pi \times 10^{-7}~~)(500)(2t)~~
	\]

	\[		<math>B(t) = 4\pi \times 10^{-4} t \ \text{T}</math>
	B(t) = ~~1000~~\pi \times 10^{-7} t
	\]

	\[		----
	~~B(t) = 4\pi \times 10^{~~-~~4} t \ \text{T}~~
	\]

	~~\hrulefill~~		'''Step 2: Take derivative of magnetic field'''

	\~~textbf~~{~~Step 2: Take derivative of magnetic field~~}		<math>\frac{dB}{dt} = 4\pi \times 10^{-4} \ \text{T/s}</math>

	\[		----
	~~\frac{dB}{dt} = 4\pi \times 10^{~~-~~4} \ \text{T/s}~~
	\]

	~~\hrulefill~~		'''Step 3: Apply Faraday’s Law'''

	\~~textbf~~{~~Step 3: Apply Faraday's Law~~}		<math>\oint \vec{E}_{nc} \cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}</math>

	\[		Magnetic flux:
	~~\oint \vec{E}_{\text{nc}} \cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}~~
	\]

	~~Magnetic flux through the solenoid area:~~		<math>\Phi_B = B \pi R^2</math>
	\[
	\Phi_B = B \pi R^2
	\]

	\[		<math>\frac{d\Phi_B}{dt} = \frac{dB}{dt} \pi R^2</math>
	\frac{d\Phi_B}{dt} = \frac{dB}{dt}\pi R^2
	\]

	Left side:		Left side of loop:
	\[
	~~\oint \vec{E}_{\text{nc}} \cdot d\ell = E_{\text{nc}} (2\pi r)~~
	\]

	\[		<math>\oint \vec{E}_{nc} \cdot d\ell = E_{nc}(2\pi r)</math>
	E_{\~~text~~{nc}} (2\pi r) ~~= - \frac{dB}{dt} \pi R^2~~
	\]

	~~Solve for the induced electric field~~:		Equate both sides:
	\[
	~~E_{\text{nc}} = -\frac{1}{2} \frac{dB}{dt} \frac{R^2}{r}~~
	\]

	\~~hrulefill~~		<math>E_{nc}(2\pi r) = -\frac{dB}{dt} \pi R^2</math>

	~~\textbf{Step 4~~: ~~Substitute numbers}~~		Solve for the induced field:

	\[		<math>E_{nc} = -\frac{1}{2}\frac{dB}{dt}\frac{R^2}{r}</math>
	E_~~{\text~~{nc}}
	= -\frac{1}{2}
	\~~left(4\pi \times 10^~~{-4}~~\right)~~
	\frac{~~(0.10)~~^2}{~~0.20~~}
	\]

	\[		----
	~~E_{\text{nc}} =~~ -~~9 \times 10^{~~-~~6} \ \text{V/m}~~
	\]

	\[		'''Step 4: Substitute numerical values'''
	~~E_{\text{nc}} = -3.14 \times 10^{-5} \ \text{V/m}~~
	\]

	(~~The negative sign indicates direction~~.)		<math>
			E_{nc} = -\frac{1}{2}(4\pi \times 10^{-4})
			\frac{(0.10)^2}{0.20}
			</math>

			<math>E_{nc} = -9 \times 10^{-6} \ \text{V/m}</math>

			Final value:

			<math>E_{nc} = -3.14 \times 10^{-5} \ \text{V/m}</math>

			(The negative sign indicates direction.)

	==Connectedness==		==Connectedness==

Pvishnubhotla6: /* Non-Coulomb Electric Field */

2025-11-30T19:22:30Z

Non-Coulomb Electric Field

@@ Line 254: / Line 254: @@
 ===Non-Coulomb Electric Field===
-Example:
+\textbf{Example:}
-A long solenoid has radius R = .10 m and contains 500 turns/meter. The current through the solenoid increases according to I(t) - 2.0t, where t is in seconds. Find the induced non-Coulomb electric field <math>/vec{E}<math> at r = .20m outside the solenoid.
 ==Connectedness==