Insulators - Revision history

Samikshasriram at 21:27, 28 September 2026

2026-09-28T21:27:29Z

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 In linear, isotropic, and homogeneous dielectric media, polarization is proportional to the electric field: ''<math>\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E}</math>''', where '''<math>\chi_e</math>''' is the electric susceptibility of the material, and '''<math>\varepsilon_0</math>''' is the vacuum permittivity. It links the electric field E and the electric displacement field D via: '''<math>\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \varepsilon_0 (1 + \chi_e) \mathbf{E} = \varepsilon \mathbf{E}</math>'''.
 ==Examples==
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 :<math>E = \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{2} - \frac{r^2}{3R} \right)</math>
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Samikshasriram: /* The Main Idea */

2026-09-28T21:26:27Z

The Main Idea

← Older revision		Revision as of 17:26, 28 September 2026
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	In linear, isotropic, and homogeneous dielectric media, polarization is proportional to the electric field: ''<math>\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E}</math>''', where '''<math>\chi_e</math>''' is the electric susceptibility of the material, and '''<math>\varepsilon_0</math>''' is the vacuum permittivity. It links the electric field E and the electric displacement field D via: '''<math>\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \varepsilon_0 (1 + \chi_e) \mathbf{E} = \varepsilon \mathbf{E}</math>'''.		In linear, isotropic, and homogeneous dielectric media, polarization is proportional to the electric field: ''<math>\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E}</math>''', where '''<math>\chi_e</math>''' is the electric susceptibility of the material, and '''<math>\varepsilon_0</math>''' is the vacuum permittivity. It links the electric field E and the electric displacement field D via: '''<math>\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \varepsilon_0 (1 + \chi_e) \mathbf{E} = \varepsilon \mathbf{E}</math>'''.

			==References==

			*https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Electricity_and_Magnetism_(Tatum)/05%3A_Capacitors/5.17%3A__Polarization_and_Susceptibility

			*https://technav.ieee.org/topic/dielectric-materials/

			*https://learnabout-electronics.org/Resistors/resistors_01a.php

			*https://web.mit.edu/6.013_book/www/chapter6/6.1.html

	==Examples==		==Examples==

Samikshasriram: /* Examples */

2026-09-28T21:23:24Z

Examples

← Older revision		Revision as of 17:23, 28 September 2026
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	===Difficult===		===Difficult===

	An infinitely long insulating cylinder of radius <math>R</math> has a non-uniform volume charge density that varies radially according to the relation <math>\rho(r) = \rho_0 \left(1 - \frac{r}{R}\right)</math> for <math>r \le R</math>, where <math>\rho_0</math> is a positive constant and <math>r</math> is the distance from the central axis. Find the magnitude of the electric field inside the cylinder (<math>r < R</math>).		An infinitely long insulating cylinder of radius <math>R</math> has a non-uniform volume charge density that varies radially according to the relation <math>\rho(r) = \rho_0 \left(1 - \frac{r}{R}\right)</math> for <math>r \le R</math>, where <math>\rho_0</math> is a positive constant and <math>r</math> is the distance from the central axis. Find the magnitude of the electric field inside the cylinder (<math>r < R</math>).

			<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
			<div style="font-weight:bold;line-height:1.6;">Solution</div>
			<div class="mw-collapsible-content">

	'''Identify the Concept:''' Because the cylinder is infinitely long and the charge depends only on <math>r</math>, the system exhibits cylindrical symmetry. We use Gauss's Law with a coaxial cylindrical Gaussian surface of radius <math>r</math> (<math>r < R</math>) and length <math>L</math>:		'''Identify the Concept:''' Because the cylinder is infinitely long and the charge depends only on <math>r</math>, the system exhibits cylindrical symmetry. We use Gauss's Law with a coaxial cylindrical Gaussian surface of radius <math>r</math> (<math>r < R</math>) and length <math>L</math>:

Samikshasriram: /* Difficult */

2026-09-28T21:23:07Z

Difficult

← Older revision		Revision as of 17:23, 28 September 2026
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	===Difficult===		===Difficult===

			An infinitely long insulating cylinder of radius <math>R</math> has a non-uniform volume charge density that varies radially according to the relation <math>\rho(r) = \rho_0 \left(1 - \frac{r}{R}\right)</math> for <math>r \le R</math>, where <math>\rho_0</math> is a positive constant and <math>r</math> is the distance from the central axis. Find the magnitude of the electric field inside the cylinder (<math>r < R</math>).


	'''Identify the Concept:''' Because the cylinder is infinitely long and the charge depends only on <math>r</math>, the system exhibits cylindrical symmetry. We use Gauss's Law with a coaxial cylindrical Gaussian surface of radius <math>r</math> (<math>r < R</math>) and length <math>L</math>:		'''Identify the Concept:''' Because the cylinder is infinitely long and the charge depends only on <math>r</math>, the system exhibits cylindrical symmetry. We use Gauss's Law with a coaxial cylindrical Gaussian surface of radius <math>r</math> (<math>r < R</math>) and length <math>L</math>:
	:<math>\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}</math>		:<math>\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}</math>

Samikshasriram: /* Examples */

2026-09-28T21:22:35Z

Examples

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 :<math>E = \frac{2Q}{4\pi \varepsilon_0 r^2} - \frac{\rho_0}{3 \varepsilon_0 r^2} (r^3 - b^3)</math>
 :<math>E = \frac{1}{4\pi \varepsilon_0} \frac{2Q}{r^2} - \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{b^3}{r^2} \right)</math>
 </div></div>

Samikshasriram: /* Intermediate */

2026-09-28T21:21:08Z

Intermediate

← Older revision		Revision as of 17:21, 28 September 2026
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	# '''Identify the Concept:''' Due to spherical symmetry, we apply Gauss's Law using a spherical Gaussian surface of radius <math>r</math> located within the insulating shell (<math>b < r < c</math>):		'''Identify the Concept:''' Due to spherical symmetry, we apply Gauss's Law using a spherical Gaussian surface of radius <math>r</math> located within the insulating shell (<math>b < r < c</math>):
	:<math>\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}</math>		:<math>\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}</math>
	:The left side evaluates to <math>E(4\pi r^2)</math> because the electric field is purely radial.		:The left side evaluates to <math>E(4\pi r^2)</math> because the electric field is purely radial.

	# '''Determine the Enclosed Charge (<math>Q_{\text{enc}}</math>):''' The total charge enclosed by our Gaussian sphere consists of two parts: the entire charge on the inner conductor, and the portion of the negative insulating shell charge located between radius <math>b</math> and radius <math>r</math>.		'''Determine the Enclosed Charge (<math>Q_{\text{enc}}</math>):''' The total charge enclosed by our Gaussian sphere consists of two parts: the entire charge on the inner conductor, and the portion of the negative insulating shell charge located between radius <math>b</math> and radius <math>r</math>.
	:<math>Q_{\text{enc}} = Q_{\text{conductor}} + Q_{\text{insulator, enc}}</math>		:<math>Q_{\text{enc}} = Q_{\text{conductor}} + Q_{\text{insulator, enc}}</math>
	:<math>Q_{\text{enc}} = 2Q + \int_b^r \rho \, dV = 2Q + (-\rho_0) \cdot \left( \frac{4}{3}\pi r^3 - \frac{4}{3}\pi b^3 \right)</math>		:<math>Q_{\text{enc}} = 2Q + \int_b^r \rho \, dV = 2Q + (-\rho_0) \cdot \left( \frac{4}{3}\pi r^3 - \frac{4}{3}\pi b^3 \right)</math>
	:<math>Q_{\text{enc}} = 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3)</math>		:<math>Q_{\text{enc}} = 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3)</math>

	# '''Apply Gauss's Law to Solve for <math>E</math>:''' Substituting <math>Q_{\text{enc}}</math> into Gauss's Law:		'''Apply Gauss's Law to Solve for <math>E</math>:''' Substituting <math>Q_{\text{enc}}</math> into Gauss's Law:
	:<math>E(4\pi r^2) = \frac{1}{\varepsilon_0} \left[ 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3) \right]</math>		:<math>E(4\pi r^2) = \frac{1}{\varepsilon_0} \left[ 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3) \right]</math>
	:Dividing both sides by <math>4\pi r^2</math> yields the electric field magnitude:		:Dividing both sides by <math>4\pi r^2</math> yields the electric field magnitude:

Samikshasriram: /* Intermediate */

2026-09-28T21:20:43Z

Intermediate

← Older revision		Revision as of 17:20, 28 September 2026
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	:<math>E = \frac{2Q}{4\pi \varepsilon_0 r^2} - \frac{\rho_0}{3 \varepsilon_0 r^2} (r^3 - b^3)</math>		:<math>E = \frac{2Q}{4\pi \varepsilon_0 r^2} - \frac{\rho_0}{3 \varepsilon_0 r^2} (r^3 - b^3)</math>
	:<math>E = \frac{1}{4\pi \varepsilon_0} \frac{2Q}{r^2} - \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{b^3}{r^2} \right)</math>		:<math>E = \frac{1}{4\pi \varepsilon_0} \frac{2Q}{r^2} - \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{b^3}{r^2} \right)</math>

	</div></div>		</div></div>

Samikshasriram: /* Intermediate */

2026-09-28T21:20:30Z

Intermediate

← Older revision		Revision as of 17:20, 28 September 2026
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	<div style="font-weight:bold;line-height:1.6;">Solution</div>		<div style="font-weight:bold;line-height:1.6;">Solution</div>
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">

			# '''Identify the Concept:''' Due to spherical symmetry, we apply Gauss's Law using a spherical Gaussian surface of radius <math>r</math> located within the insulating shell (<math>b < r < c</math>):
			:<math>\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}</math>
			:The left side evaluates to <math>E(4\pi r^2)</math> because the electric field is purely radial.

			# '''Determine the Enclosed Charge (<math>Q_{\text{enc}}</math>):''' The total charge enclosed by our Gaussian sphere consists of two parts: the entire charge on the inner conductor, and the portion of the negative insulating shell charge located between radius <math>b</math> and radius <math>r</math>.
			:<math>Q_{\text{enc}} = Q_{\text{conductor}} + Q_{\text{insulator, enc}}</math>
			:<math>Q_{\text{enc}} = 2Q + \int_b^r \rho \, dV = 2Q + (-\rho_0) \cdot \left( \frac{4}{3}\pi r^3 - \frac{4}{3}\pi b^3 \right)</math>
			:<math>Q_{\text{enc}} = 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3)</math>

	# '''Apply Gauss's Law to Solve for <math>E</math>:''' Substituting <math>Q_{\text{enc}}</math> into Gauss's Law:		# '''Apply Gauss's Law to Solve for <math>E</math>:''' Substituting <math>Q_{\text{enc}}</math> into Gauss's Law:

Samikshasriram: /* Examples */

2026-09-28T21:19:37Z

Examples

← Older revision		Revision as of 17:19, 28 September 2026
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	The magnitude of the electric field inside the insulating sphere at <math>r = 0.02 \text{ m}</math> is '''<math>4.32 \times 10^6 \text{ N/C}</math>'''.		The magnitude of the electric field inside the insulating sphere at <math>r = 0.02 \text{ m}</math> is '''<math>4.32 \times 10^6 \text{ N/C}</math>'''.


			</div></div>

			===Intermediate===
			A solid conducting sphere of radius <math>a</math> carries a net positive charge <math>2Q</math>. It is surrounded by a concentric, thick insulating spherical shell of inner radius <math>b</math> and outer radius <math>c</math> (where <math>a < b < c</math>). The insulating shell has a uniform volume charge density <math>\rho = - \rho_0</math> (a negative constant). Find the magnitude of the electric field <math>\mathbf{E}</math> in the region inside the insulator (<math>b < r < c</math>).


			<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
			<div style="font-weight:bold;line-height:1.6;">Solution</div>
			<div class="mw-collapsible-content">

			# '''Apply Gauss's Law to Solve for <math>E</math>:''' Substituting <math>Q_{\text{enc}}</math> into Gauss's Law:
			:<math>E(4\pi r^2) = \frac{1}{\varepsilon_0} \left[ 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3) \right]</math>
			:Dividing both sides by <math>4\pi r^2</math> yields the electric field magnitude:
			:<math>E = \frac{2Q}{4\pi \varepsilon_0 r^2} - \frac{\rho_0}{3 \varepsilon_0 r^2} (r^3 - b^3)</math>
			:<math>E = \frac{1}{4\pi \varepsilon_0} \frac{2Q}{r^2} - \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{b^3}{r^2} \right)</math>

	</div></div>		</div></div>

Samikshasriram: /* The Main Idea */

2026-09-28T21:18:01Z

The Main Idea

← Older revision		Revision as of 17:18, 28 September 2026
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	(Model by Samiksha Sriram)		(Model by Samiksha Sriram)

	Polarization of an insulated material occurs when molecules with pre-existing dipole moments rotate, or the dipole moments in the individual molecules are induced. This can be seen through the equation '''<math>\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}</math>''' where D is the electric displacement field, or electric flux density. This value is in the units of '''<math>\text{C/m}^2</math>''' and accounts for both free charges and the effect of polarization inside the material. ~~ε_0~~ is the permittivity of free space and is a physical constant that represents the value '''<math>8.85 \times 10^{-12} \text{ C}^2 / (\text{N} \cdot \text{m}^2)</math>'''. This value describes how electric fields behave in a vacuum. '''E''' is the electric field and is in the units N/C. Finally, '''P''' is the polarization of the medium and represents the value of the dipole moment per unit volume and is in the units '''<math>\text{C/m}^2</math>'''. Because of this equation, we know that '''<math>\mathbf{D} > \varepsilon_0 \mathbf{E}</math>''' inside the insulating material.		Polarization of an insulated material occurs when molecules with pre-existing dipole moments rotate, or the dipole moments in the individual molecules are induced. This can be seen through the equation '''<math>\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}</math>''' where D is the electric displacement field, or electric flux density. This value is in the units of '''<math>\text{C/m}^2</math>''' and accounts for both free charges and the effect of polarization inside the material. <math>\varepsilon_0</math> is the permittivity of free space and is a physical constant that represents the value '''<math>8.85 \times 10^{-12} \text{ C}^2 / (\text{N} \cdot \text{m}^2)</math>'''. This value describes how electric fields behave in a vacuum. '''E''' is the electric field and is in the units N/C. Finally, '''P''' is the polarization of the medium and represents the value of the dipole moment per unit volume and is in the units '''<math>\text{C/m}^2</math>'''. Because of this equation, we know that '''<math>\mathbf{D} > \varepsilon_0 \mathbf{E}</math>''' inside the insulating material.

	Not all materials polarize equally. Some materials are far more susceptible to polarization than others, and the ratio of the polarization of the medium to <math>\varepsilon_0 \mathbf{E}</math> is called the electric susceptibility, '''<math>\chi_e</math>''', of the medium: '''<math>\mathbf{P} = \chi_e \varepsilon_0 \mathbf{E}</math>'''. Materials that have a larger <math>\chi_e</math> value are more polarized because their charges are more easily displaced by an external electric field compared to a material with a smaller <math>\chi_e</math>.		Not all materials polarize equally. Some materials are far more susceptible to polarization than others, and the ratio of the polarization of the medium to <math>\varepsilon_0 \mathbf{E}</math> is called the electric susceptibility, '''<math>\chi_e</math>''', of the medium: '''<math>\mathbf{P} = \chi_e \varepsilon_0 \mathbf{E}</math>'''. Materials that have a larger <math>\chi_e</math> value are more polarized because their charges are more easily displaced by an external electric field compared to a material with a smaller <math>\chi_e</math>.