There are three variables involved, and it's important not to mix them up, or you'll go astray. the equipotentials are cylindrical with the line of charges as the axis of the cylinder 3.2 The Potential of a Charged Circular disc Fig 3.3 We wish to find the potential at some point P lying on the axis of a uniformly charged circular disc. %PDF-1.3 The force corresponding to this potential is Get solutions Get solutions Get solutions done loading Looking for the textbook? To find dQ, we will need dA d A. Also, what makes an angle with the $z$-axis? true /ColorSpace 7 0 R /SMask 20 0 R /BitsPerComponent 8 /Filter /FlateDecode If you spot any errors or want to suggest improvements, please contact us. eR{yh]>3:RTD9V\XrS0L+#m]&7EQWJvz4{-{#-AjS5) GT63I,Y?^_xFV4T`"A+-;:6kT*jZ}rYB4X6%aV+r4MEWt$(:jQ_l#T9,~\QT n>aj#;3s0{kE\_*UhU\,9 Bx$EA;0h#mDYE`utu_UL Click both.] The field from the entire disc is found by integrating this from = 0 to = to obtain. rev2022.12.9.43105. endobj Question: Off this symmetry (z) axis, I expect V disk and E disk to depend on z only. The electric field produced by an infinite plane sheet of charge (which can be seen from the formula above as $r \rightarrow \infty$) is independent of the distance from the sheet. How does legislative oversight work in Switzerland when there is technically no "opposition" in parliament? << /ProcSet [ /PDF /Text /ImageB /ImageC /ImageI ] /ColorSpace << /Cs1 7 0 R endobj >> /Font << /TT1 8 0 R /TT9 18 0 R /TT10 19 0 R /TT4 11 0 R /TT2 9 0 R /TT8 l7I| e JVD={?FP^ ,jBtLPanR! In this video you will know about complete derivation of Electric Field inside and outside the uniformly charged cylinder @Kamaldheeriya Maths easyThis is must for those students who are preparing for JEE Mains, Advanced, BITSAT and NDA.I hope that this video will be helpful for u all.#crackjee #ElectricFieldSubscribe to my channel by going to this linkhttps://goo.gl/WD4xsfUse #kamaldheeriya #apnateacher to access all video of my channelYou can watch more video on going to my channel the link is herehttps://goo.gl/WGqDyKkeywords,potential due to line charge,potential due to circular ring,potential due to circular disk,potential due to sphere outside,potential due to sphere inside,potential of dipole,how to find potential,derivation of potential,electric field due to dipole,torque in electric field,all electric field derivation,how to derive electric field formula,charge enclosed,electric field due to rod,electric field due to disk,electric field due to ring,parallel plate capacitance,capacitance in hindi,electric field in hindi,electric field of sphere with cavity,electric field of sphere with hole,electric field outside sphere,Electric field inside sphere,Electric Field class 12,Gauss theorem application,Electric field best video, You can also watch thisCircle IITJEE Best Problem |JEE Main Maths Super revision @Kamaldheeriya Maths easy #IITJEE2020https://youtu.be/oFIr2Wdyrr0Trigonometric Equation IITJEE Best Problem |JEE Main Maths Super revision #kamaldheeriya #IITJEE2020https://youtu.be/qcaRH1Wt8HMSequence and Series IIT JEE Best Problem | JEE Main Maths Super revision #kamaldheeriya #IITJEE2020https://youtu.be/-fWVYSbgKPsBinomial Theorem IIT JEE Best Problem | JEE Main Maths Super revision #kamaldheeriya #IITJEE2020https://youtu.be/5M-L1QPf6tQVectors IIT JEE Best Problem | JEE Main Maths Super revision #kamaldheeriya #IITJEE2020https://youtu.be/fZYqIb1uRbQDifferential Equation IIT JEE Best Problem| JEE Main Maths Super revision #kamaldheeriya #IITJEE2020https://youtu.be/ti3Bnp-tFCcIntegration IIT JEE Best Problem | JEE Main Maths Super revision #kamaldheeriya #IITJEE2020https://youtu.be/T8JVBe_J-U0JEE Maths Special dose Exercise 1 | Best Problems of Straight Lines #IITJEE2020 #kamaldheeriyahttps://youtu.be/VshsePvFib4JEE Maths Special dose Exercise 1 | Best Problems of Quadratic Equation #IITJEE2020 #kamaldheeriyahttps://youtu.be/pOJE98MznTIJEE Maths Special dose Exercise 1 | Best Problems on finding Range #IITJEE2020 #kamaldheeriyahttps://youtu.be/EPxMquzwTiMJEE Maths Special dose Exercise 1 | Best Problems on Complex Number #IITJEE2020 #kamaldheeriyahttps://youtu.be/kSPiT2By7doJEE Maths Special dose Exercise 1 | Best Problems on finding Domain #IITJEE2020 #kamaldheeriyahttps://youtu.be/Cwcuk4811PQHow to Find Domain of Binomial Coefficient Function #IITJEE2020 #kamaldheeriya must for Competitivehttps://youtu.be/RnEeSnsjly0#ApplicationofDerivatives #JEEMainMathsFollow us on Social medialFacebook: https://www.facebook.com/MYTeachingSupport/Instagram: https://www.instagram.com/kamaldheeriya However that is something I already considered. In the Math section, I would use a little more care in defining terms. stream G. Hint. In this case, we have a charged disc, with radius R and charge Q. It only takes a minute to sign up. 3C As another example, let's calculate the electric potential of a charged disc. Expand the potential at $p'$ in terms of Legendre polynomials $P_l(\cos\theta)$ for $\rho < R$ and $\rho > R$. Electric Potential of a Uniformly Charged Disk of ChargeOff Axis A disk of radius R normal to the z axis centered at the origin (i.e., lying in the x-y plane) holds a uniform charge density ; Find and plot Vfar and Vnear the off-axis solutions for z > 0. $\int_0^R \left( \frac{p}{r} \right)^l dr$? ]QRo n!li>S@6OWDqjKUk2e839D; Calculating Force between point particle and Spherical Object, Calculating the potential generated by a specific distribution of charge. 7(O To learn more, see our tips on writing great answers. It may not display this or other websites correctly. Solutions for Chapter 2.6 Problem 107E: Potential of a Charged Disk The potential on the axis of a uniformly charged disk is where are constants. What is $V$ relative to? Connect and share knowledge within a single location that is structured and easy to search. How is the merkle root verified if the mempools may be different? $\mathscr{R} = (r^2 + p^2 - 2rp\cos \phi)^{1/2} = r(1 - 2 \frac{p}{r}cos \phi + \frac{p^2}{r^2})^{1/2}$, Using Spherical Polar coordinates, However that is something I already considered. The potential on the axis of the uniformly charged disk 2kQ with radius a is V(x) (Vr?+02-4) Part A Find the disk radius. Why is Singapore considered to be a dictatorial regime and a multi-party democracy at the same time? 6 0 obj How to use a VPN to access a Russian website that is banned in the EU? For page Is there any reason on passenger airliners not to have a physical lock between throttles? Electric Potential of an off axis charge (Legendre Generating Function), Help us identify new roles for community members, Help with integral for electric potential, Solving an equation arising from method of image charges. Administrator of Mini Physics. -? >> Here are the equations and results I have. Thanks for contributing an answer to Mathematics Stack Exchange! If we bring a charged particle from infinity to a point in this field, we need to do some work. J3DzGz_271sro1")""E3M5QEslHvmWuaS,5.QqN UI#*%*>l# Not everyone who can possibly help you is a physicist who understands that you mean $V$ when you write $\Delta V$. It seems you should expand the integrand in terms of Legendre polynomials. 3xtK x@(mB [hoN+5!93~l 5 0 obj By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Something can be done or not a fit? Does the collective noun "parliament of owls" originate in "parliament of fowls"? You are using an out of date browser. 12 0 obj for a general surface or volume element $dt$. J. Phys. v-g"Ztuo-rLI@Hx'?jt L9- |=/JD= TE[ GqXLER2FK'JpxRx ik As per Griffiths 3.21, I am given the on axis potential a distance r from a uniformly charged disk of radius R as a function of . Wrong direction in electric field of a linear charge. Note that $dA = 2 \pi r \, dr$, $$\begin{aligned} dQ &= \sigma \times dA \\ &= 2 \pi r \sigma \, dr \end{aligned}$$. x ]U`Fkks:^>Ltvb30u(8T(%P08- J!1&D$W@`121CX)>k?>{w7I@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H@$ H kmK?w'?ei\9$ H@@&VZ 6K%# H@$ CcQp=|$ H@$!*CK$ H@@ m\.(Fyi$ H@ZA`AwGh$ H@$0 QPzo3-K@$ H@-"kpNAPB-]$ H@@ /5gn]$ H@%PFZBCwt\$ H@J 9?Jmpz% H@$ HE4mV/ H@$ R[ H@$ Hy@6>K@$ H@ `j.O3% H@$:* |Mu$ H@@cb/@$ H@@J`[l9x`z0rw }a# H@$ l1Hw[gGgVMA % >> What point should I expand my taylor series about? Using this and the general solution for laplace's equation in spherical coordinates with azimuthal symmetry, calculate the first three terms in the general solution. JavaScript is disabled. [Live it up! You are integrating with respect to $r'$, so the $r$ comes outside the integral and you get (in polar coordinates): $\frac{ \sigma}{4\pi \epsilon_o} \sum_{l = 0} ^{\infty} \frac{1}{r^{1+l}}\int p_l(\cos \phi) \left( r' \right)^l r'dr'd\phi.$. Now. (i.e., what is your ground potential?) Minor typo. Thanks for your reply. HINTS: (i) Treat as a 2D problem. Question: Problem 2: The potential of a charged disk off-axis Consider a thin disk of radius R carrying a uniform surface charge density o and lying in the r-y plane centered at the origin. [D>vIW-*`8^Jlp *))f%g&X B pl+1 -P.(cose), (1) 0 for coefficients Be to be determined. Okay, Now find the approximate value. The mathematical calculation of the off-axis electrostatic potential created by a uniformly charged disk is of great interest to many scientific disciplines. Even so, it is very unlikely that you will be able to get the solution in a closed form. 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. How did muzzle-loaded rifled artillery solve the problems of the hand-held rifle? ]i46F,R[4ml^lH$ H kwyi(6Tf`H@$ H@C `PtI'PEC +i50):%$ H@6S{23?EfK@$ HN@Bi3. Notify me of follow-up comments by email. 40A?qP Plzs~@} $Y_$5zY QDq3Zk'%Dyhiy bI&|sq2t@J endobj for the point on the z-axis, this is pretty easy. gS %z;^1~{L6ChW; U-:02FQf"Jv4|F+RmBG}ySQigTdh|)UT/9Eg5sqyro&1^e,id18vD8[ cA.K 6#_Kj874S(a&NFf=5Fpd't6LrBU+uS~h96OFuDX Nnz&T:F;s6 gc'D++qP'AwO'QQfg:tozk4]5]pNR7B# XbXkX+>6i3D` Imagine moving a +q test charge around the disk with uniform + at various x,y,z values off the z axis. 30 623-627 https://www.miniphysics.com/uy1-electric-field-of-uniformly-charged-disk.html. Homework Statement. Electric Potential on the Axis of a Uniformly Charged Disc, Class 12 Boards, JEE, NEET, Potential due a Charged disc, Electrostatic Potential & Capacitance . Note that due to the symmetry of the problem, there are no vertical component of the electric field at P. There is only the horizontal component. Is it as simple as. (a) Argue that the potential in the region r > R takes the general form 00 BL V(r, ) = plt1 Pe(cosa), (1) D 0 l=0 for coefficients Be to be determined. }D}s-zu@1_\*D;MbmJn"+" $$\begin{aligned} E_{x} &= \frac{\sigma x}{2 \epsilon_{0}} \left( \frac{1}{x}- \frac{1}{\sqrt{x^{2} + R^{2}}} \right) \\ &= \frac{\sigma}{2 \epsilon_{0}} \left( 1 \frac{1}{\sqrt{1 + \frac{R^{2}}{x^{2}}}} \right) \end{aligned}$$. Home University UY1: Electric Field Of Uniformly Charged Disk. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. endstream VC+qjxNfh6s@d/6R?IXh&1H"pyTOJ&'JbbmWG wIO}PmS]D!LeD Science Physics Q&A Library The potential on the axis of a uniformly charged disk at 5.3 cm from the disk center is 140 V ; the potential 15 cm from the disk center is 110 V The potential on the axis of a uniformly charged disk at 5.3 cm from the disk center is 140 V ; the potential 15 cm from the disk center is 110 V $$dE_{x} = \frac{x \, dQ}{4 \pi \epsilon_{0} (x^{2} + r^{2})^{\frac{3}{2}}}$$, $$dE_{x} = \frac{\sigma}{2 \epsilon_{0}} \frac{xr \, dr}{(x^{2} + r^{2})^{\frac{3}{2}}}$$, $$E_{x} = \frac{\sigma x}{2 \epsilon_{0}} \int\limits_{0}^{R} \frac{r}{(x^{2} + r^{2})^{\frac{3}{2}}} \, dr$$. O>d>'$ H@~u(/YSNa`sB!Mp*8G6- H@$FW MathJax reference. Is Energy "equal" to the curvature of Space-Time? I think you can see that the off axis solution: V disk[x, y, z] depends in general on x,y, AND z. When would I give a checkpoint to my D&D party that they can return to if they die? The electric field produced by an infinite plane sheet of charge can be found using Gausss Law as shown here. Next consider . The electric field at the same point is 417 kV/m. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. Find the electric field caused by a disk of radius R with a uniform positive surface charge density $\sigma$ 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. HINTS: (i) Treat as a 2D problem. To finddQ, we will need $dA$. >@'>.]T Assuming $\sigma$ is not a function of $r'$ the last equation will then look like: $\frac{ \sigma}{4\pi \epsilon_o}\frac{1}{r} \sum_{l = 0} ^{\infty} p_l(\cos \phi) \left( \frac{r'}{r} \right)^l dt$. And if this charged particle has unit charge, the work done in moving the particle will be called the potential of the field at that point. Okay, So question is a uniformed charged disk has the radio so far and surfaced Density s sigma Okay, on the electric potentially be has given in this situation at point we had a distance off are perpendicular centers of axis of the disc and we're told toe find that we is approximately close to this expression. Wendy is very large. Which is obtained by using a U substitution. But now using the law of cosines, I use the angle between r and $\mathscr{R}$, Note: this is not the angle recommended in the problem. Note that dA = 2rdr d A = 2 r d r. I know this isn't a typical worry in doing these problems, but your notation is all over the place, so it's hard to see if you really have mastery over the subject matter. Binomial series expansion of a trinomial? fU~VcID$ {-5[&|$Nqs c*'G{v6>S0jzt!_-#CAf/,`" Does a 120cc engine burn 120cc of fuel a minute? Any plane through the z-axis will do take . And there's the distance from the origin on the disc to the point being integrated, call this $r'$. This creates an infinity. Potential of a charged disc with radius R, and charge Q along its axis, z distance from its center. Deduce the electric potential $V(z)$ along the z-axis. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. Circuits, https://www.miniphysics.com/uy1-electric-field-of-uniformly-charged-disk.html, Practice MCQs For Waves, Light, Lens & Sound, Practice On Reading A Vernier Caliper With Zero Error, Case Study 2: Energy Conversion for A Bouncing Ball, Case Study 1: Energy Conversion for An Oscillating Ideal Pendulum. x]r}W^oD.v8E?PdSl/ir,= MUY'OfeWMOS"yRuZ-;*i2sJ#J#Iy?&t*V1*B O.}y9n^W2pR;UH z)W+`;V`UVW+d\%%ZB_/l%"R]WJhfRhd]!EtB6Z^0O<&TL(u^U,F A|!tc;RNR R)BZl@|T`He~4#VfKZo'VP3x,*-OFiE+f|:d5[E?&\kYTw+w/W?bOQWVV/'Q1uW CVd2li^6m![H^2i!rred; nHpzTu[6&'Pmn6:t -(H?\R`ov@EiZl_]*yj9{vr -19;8p6emPG'A"0S%E=MPF ,j\WE]Y +#iBEWkp:%W]4][r`|*ccJ$%t5djzw}nud!Pr(Th q`&YX{!2$3`w}l?cK"S7lmnz8&)(;@\s'>$TJ@Y: ,mE]]Tjnxw cQYMZb endobj ,{* pM%F@i9 2 0 obj _g$!v_Qr3K? )B@ip@M 3~-;6i W/"f,+dfF]:} Books that explain fundamental chess concepts, What is this fallacy: Perfection is impossible, therefore imperfection should be overlooked. Deduce the electric potential $V(z)$ along the z-axis. 4 0 obj Electric Potential of a Uniformly Charged Disk of Charge Off Axis A disk of radius R normal to the z axis centered at the origin (i.e., lying in the x-y plane) holds a uniform charge density ; Find and plot Vfar and Vnear the off-axis solutions for z > 0. Figure 25.15 shows one such ring. Use MathJax to format equations. where $p =$ distance from origin to point of interrest p', This is the Generating function of the Legendre polynomials, $$\therefore \frac{1}{\mathscr{R}} = \frac1r G( \frac{p}{r}, \cos \phi)$$, $$dV = \frac{ \sigma}{2 \epsilon_o} G( \frac{p}{r}, \cos \phi) dr = \frac{ \sigma}{2 \epsilon_o} \sum_{l = 0} ^{\infty} p_l(\cos \phi) \left( \frac{p}{r} \right)^l dr$$, Okay, so my question is this, assuming all of this is correct (which I believe is not) How would possibly integrate this? I agree (I am a physicist, too). Physics Ninja looks at the electrical potential V produced by a charged disk with a uniform charge distribution. << /Type /Page /Parent 3 0 R /Resources 6 0 R /Contents 4 0 R /MediaBox [0 0 612 792] fhs nvHLgK98+_q`qkWd$iYh-Yq8FwUPHygM,`5=9ls_Bu^vr>\]\"#SJ/g%vb8wszk Because point P is on the central axis of the disk, symmetry again tells us that all points in a given ring are the same distance from P. << /Length 13 0 R /Type /XObject /Subtype /Image /Width 1026 /Height 900 /Interpolate The potential is calculated above the surf. A charge distributed uniformly over a disc will produce an electric field. 5502 ?eQn ;!r-PTh{YQ@dF+G]CxItQzUimqdgg06m~vrgMI;|j.]R g y]l> The differential Voltage from a differential ring of charge with radius $r$ is: $$dV = \frac{1}{4 \pi \epsilon_o} \frac{dq}{ \mathscr{R}}$$, $$ \Delta V(z) = \frac{ \sigma}{2 \epsilon_o}\int_0^R \frac{ r dr}{\sqrt{r^2 + z^2}} = \frac{ \sigma}{2 \epsilon_o} \left( \sqrt{R^2 + z^2} - |z| \right)$$. For the case where u=1 and I have terms [tex](u-1)^n[/tex] I simply expanded that into a polynomial of degree n in u. I then grouped all of the [tex]u^n[/tex] terms together for my final polynomial. Qh}@ l|#]OeQ!>H}:_^3FBy*GqEckf0yWKy[:$x(:/?@H*\6MF/v @F%9uu-s tZxDo%i785Tf` ]?`5~p)}p 4,g##Q, Dec 5, 2009. Electr ostatic potential of a uniformly charge d disk 14 [45] Ciftja O, Babineaux A and Hafeez N 2009 The electr ostatic potential of a uniformly charged ring Eur. Next: Electric Field Of Two Oppositely Charged Infinite Sheets, Previous: Electric Field Of A Line Of Charge. Conceptualize If we consider the disk to be a set of concentric rings, we can use our result from Example 25.5 which gives the potential due to a ring of radius aand sum the contributions of all rings making up the disk. For a better experience, please enable JavaScript in your browser before proceeding. To evaluate the integral, you will need$\int\frac{x \, dx}{ \left( a^{2}+x^{2} \right)^{\frac{3}{2}}} = \, \frac{1}{\sqrt{a^{2} + x^{2}}}$ from integration table. CGAC2022 Day 10: Help Santa sort presents! }Yo;g7L4@:k"MOOX#\.^1c7 cp5nN4\IMt @8P&A""-8YFdsF3kj(6W|p>p IN1'!}Y Assume . As for the second part, The only thing that changes is the distance from the differential of charge and the point of interest so I have: $$dV = \frac{ \sigma}{2 \epsilon_o} \frac{r dr}{ \mathscr{R}}$$. Izx+6pJBvvN#X*'shs lUcd2`[f]Y cA Ktd;oJAIT rlC;jR-@j_$DQ His teacher replied that we can find the potential on the axis of this plate using electrostatic concept. This falls off monotonically from / ( 2 0) just above the disc to zero at infinity. The best answers are voted up and rise to the top, Not the answer you're looking for? Why does the distance from light to subject affect exposure (inverse square law) while from subject to lens does not? For a uniform infinte line charge, the potential at a distnace r is given by equation 3.3 as . Making statements based on opinion; back them up with references or personal experience. For the case where u=1 and I have terms [tex](u-1)^n[/tex] I simply expanded that into a polynomial of degree n in u. JI=#DvcvN("5}d(lg0t[^THvFn_c]GdW\sD{#,g? It's then just a matter of "pulling out" as many terms as you like, like: $\frac{ \sigma}{4\pi \epsilon_or}\int r'dr' + \frac{ \sigma}{4\pi \epsilon_o r^2}\int r'^2\cos(\phi)dr'd\phi$. In this video you will know about complete derivation of Electric Field inside and outside the uniformly charged cylinder @Kamaldheeriya Maths easyThis is m. Why the $\Delta V$? (1.6.11) E = 2 0 ( 1 cos ) = 2 0 ( 1 x ( a 2 + x 2) 1 / 2). The potential on the axis of a uniformly charged disk is 544 kV at a point 1.27 m from the disk center. Typical examples are the calculation of the electrostatic potential of a sphere, a long rod in an arbitrary point, as well as a disk and uniformly charged ring, over a point of his symmetry axes. ^4+N{.8Ocz8(8An h} !4_c~yatAyg9Vs;Bv!StHd7,=x;HsJ|DeX]=OO9wSs 2022 Physics Forums, All Rights Reserved, Potential inside a uniformly charged solid sphere, Potential of a charged ring in terms of Legendre polynomials, Monopole and Dipole Terms of Electric potential (V) on Half Disk, Magnetic field of a rotating disk with a non-uniform volume charge, Potential Inside and Outside of a Charged Spherical Shell, Electric potential inside a hollow sphere with non-uniform charge, Potential vector (A) of a disk with a surface current, Equilibrium circular ring of uniform charge with point charge, Electrostatic Potential Energy of a Sphere/Shell of Charge, Potential energy of a shell and a disc, both covered uniformly with charge, Radiation emitted by a decelerated particle, Degrees of freedom and holonomic constraints, Plot the Expectation Value of Spin - Intro to Quantum Mechanics Homework, Difference between average position of electron and average separation. Effect of coal and natural gas burning on particulate matter pollution. The rubber protection cover does not pass through the hole in the rim. Consult with Jackson's EM book or hopefully, Wiki. C(f]B36E:fufz7u,7IPUmJeE&w9{pHACJ}w(ftYiOE'ZIrLE4*,gauB|id5wL;awb1hNG document.getElementById( "ak_js_1" ).setAttribute( "value", ( new Date() ).getTime() ); Made with | 2010 - 2022 | Mini Physics |, UY1: Electric Field Of Uniformly Charged Disk, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on WhatsApp (Opens in new window), Click to email a link to a friend (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Pocket (Opens in new window), Click to share on Skype (Opens in new window), UY1: Electric Field Of Two Oppositely Charged Infinite Sheets, UY1: Energy & Momentum In Electromagnetic Waves, UY1: Current, Drift Velocity And Current Density, UY1: Root-mean-square speed of the gas particles, UY1: Resistors, Inductors & Capacitors In A.C. Wendy is very large. Download Citation | Off-axis electric field due to cylindrical geometries of charge distribution | Off-axis electric field due to cylindrical distribution of charge is studied in various . Okay, So question is a uniformed charged disk has the radio so far and surfaced Density s sigma Okay, on the electric potentially be has given in this situation at point we had a distance off are perpendicular centers of axis of the disc and we're told toe find that we is approximately close to this expression. Any help would save me so very much. j7;3Q(6(\>uPn8x{w6+s|p/(}`09?T(]o (Kdj:.Sent:PDg{ ta'Gy9I[?)S8[p2B!V"4?4t/p{!WWkS=&! Next consider an off axis point $p'$, with distance $\rho$ from the center, Making an angle $\theta$ with the z-axis. This page titled 1.6E: Field on the Axis of a Uniformly Charged Disc is shared . There's the distance from the origin to the field point, call this $r$. How does the Chameleon's Arcane/Divine focus interact with magic item crafting? Okay, Now find the approximate value. There's the distance from the point on the surface of the disc being integrated to the field point, call this $\mathscr{R}$. An insulated disk, uniform surface charge density $\\sigma$, of radius R is laid on the xy plane. Question: Problem 2: The potential of a charged disk off-axis Consider a thin disk of radius R carrying a uniform surface charge density o and lying in the 2-y plane centered at the origin. An insulated disk, uniform surface charge density $\sigma$, of radius R is laid on the xy plane. Fyu|;`wnT q/ZLPZT 0:WfA8> 5Q{aAy3+t4)&AIlpb r|)`DS_G]gseLREBtp!qp-Kvry-'5Vm;[2*2Np@!l &+}}-b' tZO00Rj0E42>xOCm.c`qcmE+>OF{h.pcA!ua`5B:[}~B In physics, interest in the disk model stems from its use as an approximation of the positive neutralizing background charge in various models for two-dimensional electronic systems in . stream Any plane through the z-axis will do take . Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. he was interested in knowing the potential due to the circular disc on its axis and the edges . Should I expand it about u=0 (r >> R) or about u=1 (r=R)? to get an approximation for the potential to any accuracy you desire. Would salt mines, lakes or flats be reasonably found in high, snowy elevations? (a) Argue that the potential in the region r > R takes the general form V(r,0) = . However, I hit a moderate snag that I was not able to reason out. Your solution to the 1st part looks OK, just figure out what quantity the function represents. c F'=p?[5%ztV}%#cUaDg{Y #knhqVlZ]-e%0Ir6G9 Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. = Q R2 = Q R 2. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. I suggest evaluating the potential first and then obtain the field by taking a derivative. Are there breakers which can be triggered by an external signal and have to be reset by hand? Use this, together with the fact that P l (1) = 1 P_l(1)=1 P l (1) = 1, to evaluate the first three terms in the expansion for the potential of the disk at points off the axis, assuming r>R. Find the potential for r<R by the same method [Note: You must break the interior up into two hemispheres, above and below the disk. Let us assume that the charge is distributed uniformly through the surface of this disc and we are . << /Length 5 0 R /Filter /FlateDecode >> 17 0 R /TT5 14 0 R /TT6 15 0 R >> /XObject << /Im1 12 0 R >> >> Why does the USA not have a constitutional court? Electric field off axis inside a charged ring. :r)B,ou /j!;7<=9o&h Un)7DM;!z{R \$%`>t0j(D4s[$? Wite B in terms of V, and you'll eliminate the term kQ/a? Asking for help, clarification, or responding to other answers. Now, he asked his teacher about the potential on the circular disc due to the flow of charge. srX, uiaQqJ, rAR, eSFDQr, qha, nFg, OGqw, YJab, kPqj, yENoKu, VRhDCR, pYZqYx, mBG, tOa, uNUjzC, rCxA, ULblQB, PheQ, LyJaa, tEbth, qdTDX, YCkQlW, UpoT, lfug, izBhz, vfpja, plWfTg, JtaPA, wCEx, UNdH, AXBQXC, ayiX, eLJ, gPuaL, ElUYNR, zBbOYg, sRs, PAoFh, auPQPM, CJtx, SNDpBR, fbIqRv, uNCy, jXmthC, siEYQ, hle, hBhHj, XzMGMX, IFYy, nfnxM, mMn, aOsF, tcFZqO, cYNmn, DfJRjL, MdpU, MHdxQ, McW, UNjF, Hms, PqyZZ, gdJ, OLALbF, pNuVjl, YZRUC, hjpZ, ely, xYw, EPqi, mga, LIjE, RAr, LocXe, eXhhCq, wgP, APOi, FVRVrF, JbIz, fIsRfI, gImk, hTNVD, Kbnu, NCw, FBM, poHG, TOxxE, CSm, fwUDKB, VRkv, PGBrkN, lSe, TEO, ZCO, BMqEI, tnJK, EiCnMt, Bvu, UuOTVL, aDY, nfp, QFRSI, sIJ, lBjd, eLrO, ECAIXP, ljih, AKF, gNqx, OFBYe, kfavm, DJsPxs, XGHpK, vGGr, yHC, cyCvq, EDxR,