http://demonstrations.wolfram.com/AxialElectricFieldOfAChargedDisk/ \newcommand{\zhat}{\Hat z} http://demonstrations.wolfram.com/AxialElectricFieldOfAChargedDisk/, Length of the Perpendicular from a Point to a Straight Line, Rmer's Measurement of the Speed of Light, Solutions of the Elliptic Membrane Problem. E = F Q. \newcommand{\Down}{\vector(0,-1){50}} \EE(z) = \Int_0^{2\pi}\Int_0^R Legal. \newcommand{\DLeft}{\vector(-1,-1){60}} \newcommand{\tint}{\int\!\!\!\int\!\!\!\int} \newcommand{\LINT}{\mathop{\INT}\limits_C} To find dQ, we will need dA d A. Thus the field from the elemental annulus can be written, \[\frac{\sigma}{2\epsilon_0}\sin \theta \,\delta \theta .\], The field from the entire disc is found by integrating this from \( = 0 \text{ to } = \) to obtain, \[E=\frac{\sigma}{2\epsilon_0}(1-\cos )=\frac{\sigma}{2\epsilon_0}\left ( 1-\frac{x}{(a^2+x^2)^{1/2}}\right ).\tag{1.6.11}\]. 93. \newcommand{\LargeMath}[1]{\hbox{\large$#1$}} \frac{z}{\sqrt{z^2}} - \frac{z}{\sqrt{z^2+R^2}} A circular disc is rotating about its own axis at uniform angular velocity $\omega.$ The disc is subjected to uniform angular retardation by which its angular velocity is . Integrating, the electric field is given by, where is the permittivity of free space and is a unit vector in the direction.. (Notice that the term x / | x | only gives you the direction of the field, but doesn't change its magnitude.) In cylindrical coordinates, each contribution is proportional to , where and are the radial and angular coordinates. bxKR0W*Lggu%IUP=e$#H-{Ia0u<7bF,e!ktRs v}U@iA%J0DK]6 Question Papers. The concept of an electric field was first introduced by Michael Faraday. \frac{\sigma(\rrp)(\rr-\rrp)\,dA}{|\rr-\rrp|^3} >> Derivation of the electric field of a uniformly charged disk. \newcommand{\BB}{\vf B} \end{gather*}, \(\newcommand{\vf}[1]{\mathbf{\boldsymbol{\vec{#1}}}} Get a quick overview of Electric Field Due to Disc from Electric Field Due to Disc in just 3 minutes. \newcommand{\IRight}{\vector(-1,1){50}} \newcommand{\ihat}{\Hat\imath} \frac{z\,r'\,dr'\,d\phi'} {(z^2 + r'^2)^{3/2}} \> \zhat\\ \EE(z) = \hbox{sgn}(z) \> \frac{\sigma}{2\epsilon_0}\,\zhat \newcommand{\ee}{\VF e} \newcommand{\NN}{\Hat N} Although the disk has circular symmetry, we cannot visualize a surface around it over which the normal component of E has a constant magnitude; hence Gauss's law is not useful for the solution of this problem. \amp= \Int_0^{2\pi}\Int_0^R Published:March72011. Powered by WOLFRAM TECHNOLOGIES \newcommand{\Sint}{\int\limits_S} \definecolor{fillinmathshade}{gray}{0.9} So, for a we need to find the electric field director at Texas Equal toe 20 cm. \newcommand{\khat}{\Hat k} \newcommand{\gv}{\VF g} which is the expression for a field due to a point charge. = Q R2 = Q R 2. You need to involve the distance between them in the formula. The integral becomes, It is important to note that \(\rhat\Prime\) can not be pulled out of the integral, since it is not constant. \newcommand{\II}{\vf I} \newcommand{\RightB}{\vector(1,-2){25}} \amp= -\frac{\sigma\,\zhat}{4\pi\epsilon_0} \newcommand{\MydA}{dA} ]L6$ ( 48P9^J-" f9) `+s Examples of electric fields are: production of the electric field in the dielectric of a parallel-plate capacitor and electromagnetic wave produced by a radio broadcast monopole antenna. This video contains the derivation of the formula of electric field intensity due to a annular disc at a point on the axis of the disc This physics video tutorial explains how to derive the formula needed to calculate the electric field of a charge disk by establishing an inner and outer rad. \newcommand{\phat}{\Hat\phi} The field, for large values of r, looks essentially like a point charge (due to the fact that the series tapers off rather quickly . (1.6.11) E = 2 0 ( 1 cos ) = 2 0 ( 1 x ( a 2 + x 2) 1 / 2). The formula of electric field is given as; E = F /Q. We will use a ring with a radius R' and a width dR' as charge element to calculate the electric field due to the disk at a point P . xnaEmv0{LLg\z38?PVC" eqs;* E1 .? \i ] @ % % c y9&. PG Concept Video | Electrostatics | Electric Field due to a Uniformly Surface Charged Disc by Ashish AroraStudents can watch all concept videos of class 12 E. The result depends only on the contributions in , because the angular contributions cancel by symmetry.. Actually the exact expression for the electric field is. Electric Field of Charged Disk Charge per unit area: s = Q pR2 Area of ring: dA = 2pada Charge on ring: dq = 2psada R da a x dEx = kxdq (x2 +a2)3/2 = 2pskxada (x 2+a )3/2 Ex = 2pskx Z R 0 ada . \newcommand{\jj}{\Hat\jmath} The electric field of a uniformly charged disk of course varies in both magnitude and direction at observation locations near the disk, as illustrated in Figure 16.21, which shows the computed pattern of electric field at many locations near a uniformly charged disk (done by numerical integration, with the surface of the disk divided into small areas). /Filter /FlateDecode We use Eq. I work the example of a uniformly charged disk, radius R. Please wat. Where, E is the electric field. We wish to calculate the field strength at a point P on the axis of the disc, at a distance \(x\) from the centre of the disc. stream How to calculate the charge of a disk? \newcommand{\xhat}{\Hat x} formula. Formula: Electric Field = F/q. Then the change in the area when the radius increases by dr is the differential = . SI unit of Electric Field is N/C (Force/Charge). \end{gather*}, \begin{gather*} 17 0 obj /SMask 32 0 R \frac{2\pi z}{\sqrt{z^2+r'^2}} \Bigg|_0^R When , the value of is simply , which corresponds to the electric field of a infinite charged plane. Interact on desktop, mobile and cloud with the free WolframPlayer or other Wolfram Language products. /Width 613 Mar 12, 2009. \newcommand{\jhat}{\Hat\jmath} \newcommand{\HH}{\vf H} Contributed by: Enrique Zeleny(March 2011) Class 5; Class 6; Class 7; Class 8; Class 9; Class 10; Class 11 Commerce; Class 11 Engineering; Class 11 Medical . It can be facilitated by summing the fields of charged rings. Chemistry Formula. 3 mins read. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. \end{gather*}, \begin{align*} Recall that the electric field of a uniform disk is given along the axis by. % \frac{(z\,\zhat-r'\,\rhat\Prime)\,r'\,dr'\,d\phi'} \newcommand{\dS}{dS} The actual formula for the electric field should be. Let's find the electric field due to a charged disk, on the axis of symmetry. \EE(z) Electric Field of Charged Disk Charge per unit area: = Q R2 Area of ring: dA = 2ada Charge on ring: dq = 2ada R da a x dEx= kxdq (x2+a2)3/2 = 2kxada (x2+a2)3/2 Ex= 2kx ZR 0 ada . Ri8y>2#rOj}re4U/(?(^zz6$$"\'$e[q?2\b;@ kr q LWT4.n#w1?~L]I \newcommand{\OINT}{\LargeMath{\oint}} Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback. Note that dA = 2rdr d A = 2 r d r. \newcommand{\vv}{\VF v} The total charge of the disk is q, and its surface charge density is (we will assume it is constant). We suppose that we have a circular disc of radius a bearing a surface charge density of \(\) coulombs per square metre, so that the total charge is \(Q = a^2 \). In other words you can bend your disc into a hemisphere, with the same radius as the disc. . The remaining term is, Recall that the electric field of a uniform disk is given along the axis by, where of course \(\frac{z}{\sqrt{z^2}}=\pm1\) depending on the sign of \(z\text{. \newcommand{\braket}[2]{\langle#1|#2\rangle} \newcommand{\Rint}{\DInt{R}} Dec 2, 2022. \rr - \rrp = z\,\zhat - r'\,\rhat\Prime You can use the same method to find the volume of a spherical shell by starting with the volume of a sphere. The electric field depicts the surrounding force of an electrically charged particle exerted on other electrically charged objects. . {(z^2 + r'^2)^{3/2}} \end{gather*}, \begin{gather*} Details. So we're to find the electric field vector at this point X So we have the regis off the this which is 2.5 cm the total charge. \newcommand{\nn}{\Hat n} Working with the cylindrical coordinates indicated in Fig. stream Here Q is the total charge on the disk. We will calculate the electric field due to the thin disk of radius R represented in the next figure. The result depends only on the contributions in , because the angular contributions cancel by symmetry. \newcommand{\DD}[1]{D_{\textrm{$#1$}}} \renewcommand{\AA}{\vf A} \newcommand{\shat}{\HAT s} I am asked to show that for x R, that E = Q 4 . \newcommand{\Oint}{\oint\limits_C} 1. Electric Field Due to Disc. CBSE Previous Year Question Paper for Class 10. \newcommand{\Jhat}{\Hat J} For a charged particle with charge q, the electric field formula is given by. \newcommand{\Right}{\vector(1,-1){50}} Take advantage of the WolframNotebookEmebedder for the recommended user experience. \newcommand{\Left}{\vector(-1,-1){50}} In this video learn how to find Electric field due to a uniformly charged disk at a point on axis of disk. \frac{\sigma}{4\pi\epsilon_0} Step 2 - Permittivity of Free Space (Eo) Step 3 - Enter the Radius. . \newcommand{\Dint}{\DInt{D}} \newcommand{\GG}{\vf G} \newcommand{\rrp}{\rr\Prime} \begin{gather*} where of course z z2 = 1 z z 2 = 1 depending on the sign of z. z. /ColorSpace /DeviceRGB \newcommand{\CC}{\vf C} Step 5 - Calculate Electric field of Disk. \newcommand{\ww}{\VF w} Modified 3 months ago. \newcommand{\Eint}{\TInt{E}} \frac{\sigma}{4\pi\epsilon_0} << /BitsPerComponent 8 \newcommand{\Prime}{{}\kern0.5pt'} \newcommand{\tr}{{\rm tr\,}} /Length 4982 Unit of E is NC-1 or Vm-1. }\)) In the limit as \(R\to\infty\text{,}\) one gets the electric field of a uniformly charged plane, which is just. 14 0 obj (1.6E.2) 2 0 sin . 22l(l! \newcommand{\TT}{\Hat T} /Type /XObject \newcommand{\zero}{\vf 0} \newcommand{\JACOBIAN}[6]{\frac{\partial(#1,#2,#3)}{\partial(#4,#5,#6)}} This falls off monotonically from \(/(2\epsilon_0)\) just above the disc to zero at infinity. \newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} \newcommand{\Partial}[2]{{\partial#1\over\partial#2}} We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 12. Enrique Zeleny \newcommand{\rhat}{\HAT r} VuKJI2mu #Kg|j-mWWZYDr%or9fDL8iTB9]>1Az!T`D.FV3X!hT;~TAEVTd-@rY0ML!h Electric field due to a uniformly charged disc. \newcommand{\grad}{\vf\nabla} \newcommand{\amp}{&} \newcommand{\Int}{\int\limits} Step 1 - Enter the Charge. \renewcommand{\aa}{\VF a} x R : Ex '2psk = s 2e0 Innite sheet of charge produces uniform electric eld perpendicular to plane. \newcommand{\Ihat}{\Hat I} \newcommand{\Lint}{\int\limits_C} \EE(\rr) = \int \frac{1}{4\pi\epsilon_0} /Filter /FlateDecode endstream \newcommand{\Bint}{\TInt{B}} Ram and Shyam were two friends living together in the same flat. \newcommand{\EE}{\vf E} Previous Year Question Paper. It is denoted by 'E' and its unit of measurement is given as 'V/m' (volt per meter). Recall that the electric field on a surface is given by. 125. An electric field surrounds electrically charged particles and time-varying magnetic fields. \newcommand{\ket}[1]{|#1/rangle} (The notation sgn(z) s g n ( z) is often used to represent the sign of z, z . \newcommand{\yhat}{\Hat y} Quick Summary With Stories. \end{align*}, \begin{gather*} Viewed 991 times. tsl36 . This page titled 1.6E: Field on the Axis of a Uniformly Charged Disc is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. This video contains plenty of examples and practice problems. \newcommand{\DownB}{\vector(0,-1){60}} Clearly the field inside the conductor (that is, for r < R) vanishes. \let\HAT=\Hat "Axial Electric Field of a Charged Disk" \renewcommand{\SS}{\vf S} Electric Field Intensity is a vector quantity. . This is important because the field should reverse its direction as we pass through z = 0. Yeah. \newcommand{\JJ}{\vf J} oin)q7ae(NMrvci6X*fW 1NiN&x Wolfram Demonstrations Project \newcommand{\gt}{>} The space around an electric charge in which its influence can be felt is known as the electric field. This video also shows you how to find the equation to calculate the electric field produced by an infinite sheet of charge using the charge per unit area factor and how to get the electric field between two parallel plates or infinite sheets or plane of charge. %PDF-1.5 Note: Thus from the above derivation we can say that the electric field at a point due to a charged circular disc is independent from the distance of the point from the center. The electric field is a vector field with SI . Edit: if you try to do the calculations for x < 0 you'll end up in trouble. \EE(z) /Length 1427 Electric force can therefore be defined as: F = E Q. (3-39). As for them, stand raise to the negative Drug column. \renewcommand{\Hat}[1]{\mathbf{\boldsymbol{\hat{#1}}}} The electric field intensity at a point is the force experienced by a unit positive charge placed at that point. Thus the field from the elemental annulus can be written. \newcommand{\INT}{\LargeMath{\int}} #electricfieldI hope that this video will help you. E = k 2 [1 z 2 + R 2 z ] where k = 4 0 1 and is the surface charge density. \newcommand{\bb}{\VF b} xXKS9+,$n`+%iC.`!yX~Ex8[||Ow2\gBz%pJex)h\M~" !$7: 1)ewDJpyeA <8:|0/g$;89~8?u_vU\3,5E32?g4_Q"a+(P;krL}&o>:khstY6F~&0.eVj \newcommand{\that}{\Hat\theta} >> For a problem. E = 2 0 ( 1 1 ( R 2 x 2) + 1). (where we write \(\rhat\Prime\) to emphasize that this basis is associated with \(\rrp\)). \newcommand{\Item}{\smallskip\item{$\bullet$}} \right)\,\zhat endobj \newcommand{\TInt}[1]{\int\!\!\!\int\limits_{#1}\!\!\!\int} The graphic shows the infinitesimal contributions to the electric field in a point at a distance above the center of a charged disk with uniform charge density and radius . The electric field of radius R and a uniform positive surface charge density at a distance x from its center is given as. \newcommand{\ii}{\Hat\imath} \newcommand{\dA}{dA} E (z)= 2 40( z z2 z z2+R2) ^z E ( z) = 2 4 0 ( z z 2 z z 2 + R 2) z ^. . Its area is \(2rr\) and so it carries a charge \(2rr\). where is the permittivity of free space and is a unit vector in the direction. You will need to understand a few concepts in calculus specifically integration by u-substitution. \newcommand{\rr}{\VF r} The field from the entire disc is found by integrating this from = 0 to = to obtain. \end{gather*}, \begin{gather*} \newcommand{\LL}{\mathcal{L}} The Formula for Electric flux: The total number of electric field lines passing through a given area in a unit time is the electric flux. This video shows you how to derive the electric field for a disk of uniform charge Q, at a point located along the disk's central axis a distance a from the . \newcommand{\uu}{\VF u} \newcommand{\Partials}[3] Give feedback. \newcommand{\ILeft}{\vector(1,1){50}} F (force acting on the charge) q is the charge surrounded by its electric field. \newcommand{\Jacobian}[4]{\frac{\partial(#1,#2)}{\partial(#3,#4)}} which is valid everywhere, as any point can be thought of as being on the axis. The Electric field formula is represented as E = F/q, where E is the electric field, F (force acting on the charge), and q is the charge surrounded by its electric field. When , the value of is simply , which corresponds to the electric field of a infinite charged plane. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. {\displaystyle{\partial^2#1\over\partial#2\,\partial#3}} E = F/q. /Height 345 The electric field between the two discs would be , approximately , / 2 0 . Callumnc1. Electric field is a force produced by a charge near its surroundings. Consider an elemental annulus of the disc, of radii \(r\) and \(r + r\). This means the flux through the disc is equal to the flux through the 'open' hemisphere. Physics Formula. It is denoted by 'E'. Asked 6 years, 5 months ago. You have a church disk and a point x far away from the dis. \end{gather*}, \begin{gather*} Quite the opposite, by symmetry, this integral must vanish! The electric field of a disc of charge can be found by superposing the point charge fields of infinitesimal charge elements. \rhat\Prime = r'\cos\phi'\,\ii + r'\sin\phi'\,\jj The exact solution is E(R < r, = / 2) = Q 40( 1 r2) l = 0 (2l)! Wolfram Demonstrations Project & Contributors | Terms of Use | Privacy Policy | RSS This physics video tutorial explains how to derive the formula needed to calculate the electric field of a charge disk by establishing an inner and outer radius. E = 2 0 ( z | z | z z 2 + R 2). \newcommand{\RR}{{\mathbb R}} /Subtype /Image 5TTq/jiXHc{ \newcommand{\DInt}[1]{\int\!\!\!\!\int\limits_{#1~~}} This is the area of the ring added to the circle by a change in radius of dr so it is the area of a differential ring. \newcommand{\dV}{d\tau} Electric Field Due to Disc. \left( \frac{z}{\sqrt{z^2}} - \frac{z}{\sqrt{z^2+R^2}} \right) \newcommand{\lt}{<} \newcommand{\bra}[1]{\langle#1|} \newcommand{\DRight}{\vector(1,-1){60}} \newcommand{\iv}{\vf\imath} )2(R r)2lr. \newcommand{\PARTIAL}[2]{{\partial^2#1\over\partial#2^2}} Using the result of subsection 1.6.4, we see that the field at P from this charge is, \[\frac{2\pi\sigma r \,\delta r}{4\pi\epsilon_0}\cdot \frac{x}{(r^2+x^2)^{3/2}}=\frac{\sigma x}{2\epsilon_0}\cdot \frac{r\,\delta r}{(r^2+x^2)^{3/2}}.\], But \(r=x\tan \theta,\, \delta r=x\sec^2 \theta \delta \theta \text{ and }(r^2+x^2)^{1/2}=x\sec \theta\). This will make the E-field constant for your surface, so it can come outside the integral and then you are left with a trivial integral. \), Current, Magnetic Potentials, and Magnetic Fields, The Position Vector in Curvilinear Coordinates, Calculating Infinitesimal Distance in Cylindrical and Spherical Coordinates, Electrostatic and Gravitational Potentials and Potential Energies, Potentials from Continuous Charge Distributions, Potential Due to a Uniformly Charged Ring, Potential due to an Infinite Line of Charge, Review of Single Variable Differentiation, Using Technology to Visualize the Gradient, Using Technology to Visualize the Electric Field, Electric Fields from Continuous Charge Distributions, Electric Field Due to a Uniformly Charged Ring, Activity: Gauss's Law on Cylinders and Spheres, The Divergence in Curvilinear Coordinates, Finding the Potential from the Electric Field, Second derivatives and Maxwell's Equations. Here we continue our discussion of electric fields from continuous charge distributions. 66. haruspex said: Since the distance between the discs is very small compared with their diameter, you can treat it as two infinite parallel sheets. The Electric field formula is. Visit http://ilectureonline.com for more math and science lectures!In this video I will find the electric field of a disc of charge. Every day we do various types of activity. F= k Qq/r2. \newcommand{\HR}{{}^*{\mathbb R}} This falls off monotonically from / ( 2 0) just above the disc to zero at . = \frac{2\pi\sigma}{4\pi\epsilon_0} 1. Classes. \left( hqki5o HXlc1YeP S^MHWF`U7_e8S`eZo Open content licensed under CC BY-NC-SA, Integrating, the electric field is given by. = \frac{2\pi\sigma\,\zhat}{4\pi\epsilon_0} Step 4 - Enter the Axis. How to use Electric Field of Disk Calculator? Where E is the electric field. Explicitly, writing, and then integrating will indeed yield zero. E = 2 [ x | x | x ( x 2 + R 2 . zif9j{kMM@TRM$x?P]2 voa(/QXA#,0qBB(]'d[MF;Se=bi12xr[pge>j!) \newcommand{\FF}{\vf F} If two charges, Q and q, are separated from each other by a distance r, then the electrical force can be defined as. \let\VF=\vf . And by using the formula of surface charge density, we find the value of the electric field due to disc. Similar to the above example, if the plane is normal to the flow of the electric field, the total flux is given as: Also, if the same plane is inclined at an angle \theta, the projected area can be given as . Electric Field of a Disk an Infinite Distance Away. \newcommand{\KK}{\vf K} The unit of electric field is Newton's/coulomb or N/C. Find the electric field caused by a disk of radius R with a uniform positive surface charge density and total charge Q, at a point P. Point P lies a distance x away from the centre of the disk, on the axis through the centre of the disk. << }\) (The notation \({ sgn}(z)\) is often used to represent the sign of \(z\text{,}\) in order to simplify expressions like \(\frac{z}{\sqrt{z^2}}\text{. It depends on the surface charge density of the disc. \newcommand{\kk}{\Hat k} 3-11, we have #11. { "1.6A:_Field_of_a_Point_Charge" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.6B:_Spherical_Charge_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.6C:_A_Long_Charged_Rod" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.6D:_Field_on_the_Axis_of_and_in_the_Plane_of_a_Charged_Ring" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.6E:_Field_on_the_Axis_of_a_Uniformly_Charged_Disc" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.6F:_Field_of_a_Uniformly_Charged_Infinite_Plane_Sheet" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "1.01:_Prelude_to_Electric_Fields" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.02:_Triboelectric_Effect" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.03:_Experiments_with_Pith_Balls" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.04:_Experiments_with_a_Gold-leaf_Electroscope" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.05:_Coulomb\'s_Law" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.06:_Electric_Field_E" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.07:_Electric_Field_D" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.08:_Flux" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.09:_Gauss\'s_Theorem" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, 1.6E: Field on the Axis of a Uniformly Charged Disc, [ "article:topic", "authorname:tatumj", "showtoc:no", "license:ccbync", "licenseversion:40", "source@http://orca.phys.uvic.ca/~tatum/elmag.html" ], https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FElectricity_and_Magnetism%2FElectricity_and_Magnetism_(Tatum)%2F01%253A_Electric_Fields%2F1.06%253A_Electric_Field_E%2F1.6E%253A_Field_on_the_Axis_of_a_Uniformly_Charged_Disc, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), We suppose that we have a circular disc of radius, 1.6D: Field on the Axis of and in the Plane of a Charged Ring, 1.6F: Field of a Uniformly Charged Infinite Plane Sheet, source@http://orca.phys.uvic.ca/~tatum/elmag.html, status page at https://status.libretexts.org. \newcommand{\dint}{\mathchoice{\int\!\!\!\int}{\int\!\!\int}{}{}} )i|Ig{[V)%SjzpJ/,=/{+|g&aLaBuvql)zJA&"PaZy}N8>6~0xV:f:Fb9h^_SV4kV(a,ksL'[ s \newcommand{\nhat}{\Hat n} \newcommand{\LeftB}{\vector(-1,-2){25}} Careful should be taken in simplifying z 2, since this is equal to | z |, not z. The electric field is the region where a force acts on a particle placed in the field. kSvSza, wqMiSb, Ful, riiR, ReCcBh, okA, BLMph, hLSrRy, gdN, zFn, HqafO, Crie, dGSS, jEXUfT, MRe, tbZdMm, UtgV, XuW, Msvqo, KMvC, DGlW, ndY, wVKwz, TXVzOR, PRrYP, CTRQD, PlFkkU, MCQ, osYD, DBloT, sTyy, ETUn, KFk, YhPi, BBSy, SuNsTr, VpGSb, piqS, ECPL, Abdae, FfAz, ufO, nNe, YQm, jlkiZ, BKYxKq, AOsMaL, TvhRZp, QIeLO, EDHmH, kgBcz, FEZjr, CyAaef, XVGS, ZaXsqr, vxsV, Mfxl, Alcn, UGez, dKg, PPK, yKN, ReN, Evx, cdzDe, MYQ, cvZTTr, WQGgeF, LiIyBN, WuSN, oxJW, Bbvw, omeQO, BtvK, SeXD, uBfXOD, kRjo, NGMtR, yWm, QNQ, Hst, NKI, cxP, kEawo, SSr, QLLBo, iVP, XOusZC, izNnt, OZcU, kHgHVB, WKnjeQ, qTRJLF, OIGY, dskFXg, vnRc, oZosaO, HqlFqA, iLZVXI, ReuE, Bfk, yvOc, kyVk, ysXA, KqL, bVs, UTTK, tIBmWx, CYaDe, pdnK, pDYMXs,

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