What about the octopole term? \renewcommand{\SS}{\vf S} You are using an out of date browser. Let the linear charge density of this wire be . P is the point that is located at a perpendicular distance from the wire. dE = (Q/Lx2)dx 40 d E = ( Q / L x 2) d x 4 0. For a point charge, the potential V is related to the distance r from the charge q, V = 1 4 0 q r. This is the question I have: consider the system formed by two infinitely long line charges located in the xy plane running parallel to the x axis at y = + and - a and carrying uniform charge densities + and - lambda respectively. The pontential difference increases as you go farther. Attempt : I first calculate the potential for x = 0. See Answer. Think about it graphically. Now we wish to calculate the potential at point r from the linear charge distribution. dq = Q L dx d q = Q L d x. Is +ln(r) or -ln(r) sloping up? \newcommand{\rhat}{\HAT r} Get access to the latest Electric Potential Due to Disc and Infinite line of charge. Now, we want to plug in information, using the use what you know strategy. Scalar Line Integrals; Vector Line Integrals; General Surface . \newcommand{\ket}[1]{|#1/rangle} \newcommand{\LargeMath}[1]{\hbox{\large$#1$}} \newcommand{\DownB}{\vector(0,-1){60}} \newcommand{\ee}{\VF e} Making statements based on opinion; back them up with references or personal experience. \newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} Potential due to an Infinite Line of Charge 9 Differentials Review of Single Variable Differentiation Leibniz vs. Newton Differentials The Multivariable Differential Rules for Differentials Properties of Differentials Differentials: Summary 10 Gradient The Geometry of Gradient The Gradient in Rectangular Coordinates Properties of the Gradient Electric Potential Due to Continuous Charge Distributions A rod of length L (Figure) lies along the x axis with its left end at the origin. From the above potential formula, we have $\eqalign{ & E = - \dfrac{{dV}}{{dx}} \cr The potential at all the points on the axis will be zero. When calculating the difference in electric potential due with the following equations. Dealing with a point charge is very easy and convenient as the electric field is originated from a point source. \newcommand{\INT}{\LargeMath{\int}} ($\dfrac{dt}{t} dt$ shouldnt this just be $\dfrac{dt}{t}$? {\displaystyle{\partial^2#1\over\partial#2\,\partial#3}} \newcommand{\ww}{\VF w} Lesson 16 of 27 5 upvotes 12:58mins. It has a nonuniform charge density = x, where a is a positive constant. \newcommand{\jhat}{\Hat\jmath} Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. \newcommand{\KK}{\vf K} \newcommand{\FF}{\vf F} \newcommand{\PARTIAL}[2]{{\partial^2#1\over\partial#2^2}} The direction of the electric field at all points on the axis will be along the axis If the ring is placed inside a uniform external electric field, then net torque and force acting on the ring would be zero. This is the question I have: consider the system formed by two infinitely long line charges located in the xy plane running parallel to the x-axis at y = + and - a and carrying uniform charge densities + and - lambda respectively. \newcommand{\nn}{\Hat n} What answer do you expect? Effect of coal and natural gas burning on particulate matter pollution. \newcommand{\HR}{{}^*{\mathbb R}} \begin{equation} You're using cylindrical coordinates (because of the symmetry of the problem), and you integrate along $r$, which is $|\vec{r}|$. What confuses me is that the $\ln()$ is negative. Is the EU Border Guard Agency able to tell Russian passports issued in Ukraine or Georgia from the legitimate ones? \renewcommand{\Hat}[1]{\mathbf{\boldsymbol{\hat{#1}}}} \newcommand{\LINT}{\mathop{\INT}\limits_C} JavaScript is disabled. Electric field due to infinite line charge: Bounds of integration? Why is Singapore considered to be a dictatorial regime and a multi-party democracy at the same time? electric-fields electrostatics homework-and-exercises potential. \newcommand{\GG}{\vf G} Find the elctrical potential at all points in space using the origin as your referenc point. \newcommand{\HH}{\vf H} \newcommand{\yhat}{\Hat y} So where is the error? ##\vec{E}## in the integrand is a vector. CGAC2022 Day 10: Help Santa sort presents! \newcommand{\vv}{\VF v} \definecolor{fillinmathshade}{gray}{0.9} JavaScript is disabled. For that let us use equation 1 to determine the potential at point r with respect to the reference point ${{r}_{o}}$. \newcommand{\Jhat}{\Hat J} Because potential is defined with respect to infinity. \newcommand{\ihat}{\Hat\imath} View Solution Q. No tracking or performance measurement cookies were served with this page. To elaborate a bit on Bill's comment, you might consider a curve defined as follows in some cylindrical $(r,\theta,z)$ coordinate system: $$\gamma(t) = \big(r(t),\theta(t),z(t)\big) = (t, 0, 0)$$ $$\gamma(t) = \big(r(t),\theta(t),z(t)\big) = (t, 0, 0)$$, Am I missing something, or you have an extra $dt$ on your 4th equality? Is there really no meaning in potential energy and potential? For infinite length . 2022 Physics Forums, All Rights Reserved, Electromagnetic linear momentum for a system of two moving charges, Potential of a charged ring in terms of Legendre polynomials, Electrostatic Potential Energy of a Sphere/Shell of Charge, Time needed to cross a delta potential barrier inside an infinite square well, The potential of a sphere with opposite hemisphere charge densities, Potential Inside and Outside of a Charged Spherical Shell, Exponential Wavefunction for Infinite Potential Well Problem, Potential outside a grounded conductor with point charge inside, Potential at the origin due to an infinite set of point charges, Equilibrium circular ring of uniform charge with point 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. Best Answer. \newcommand{\ii}{\Hat\imath} \newcommand{\tr}{{\rm tr\,}} \newcommand{\Eint}{\TInt{E}} \newcommand{\IRight}{\vector(-1,1){50}} MathJax reference. \newcommand{\Ihat}{\Hat I} \newcommand{\TInt}[1]{\int\!\!\!\int\limits_{#1}\!\!\!\int} What is the length of an infinite potential well for an electron? [Physics] Potential due to line charge. E=dV/dr is always positive, so V should be sloping up. Share Cite Improve this answer Follow edited May 23, 2018 at 19:13 answered May 23, 2018 at 19:08 V.F. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators . Problem: Two infinite planes with surface charges + and are perpendicular to the x -axis at x = 0 and x = 2 respectively. We will choose to work in cylindrical coordinates, centering the line segment on the \(z\)-axis and will find the potential at a distance \(s\) from the origin in the \(x\text{,}\) \(y\)-plane, as shown in Figure8.7.1. }\) Because we are chopping a one-dimensional source into little lengths, \(d\tau\) reduces to \(|d\rr|\text{.}\). \int_{-L}^{L}\frac{dz'}{\sqrt{s^2+z'^2}}\\ Asking for help, clarification, or responding to other answers. V = 40 ln( a2 + r2 +a a2 + r2-a) V = 4 0 ln ( a 2 + r 2 + a a 2 + r 2 - a) We shall use the expression above and observe what happens as a goes to infinity. V(\rr) = \frac{1}{4\pi\epsilon_0}\int\frac{\lambda |d\rr|}{|\rr-\rrp|} .\tag{8.7.1} To subscribe to this RSS feed, copy and paste this URL into your RSS reader. \let\VF=\vf You are using an out of date browser. \newcommand{\Sint}{\int\limits_S} \amp= \frac{\lambda}{4\pi\epsilon_0} \newcommand{\Jacobian}[4]{\frac{\partial(#1,#2)}{\partial(#3,#4)}} \newcommand{\xhat}{\Hat x} 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. \newcommand{\shat}{\HAT s} \newcommand{\Prime}{{}\kern0.5pt'} \amp= \frac{\lambda}{4\pi\epsilon_0} 6 Potentials due to Discrete Sources. That's because kdq/r assumes you're taking V = 0 at infinity. \newcommand{\CC}{\vf C} }\) Alternatively, this result can be obtained directly from a diagram using the Pythagorean Theorem. The less you move away, the more similar potential you have (little difference). Does integrating PDOS give total charge of a system? Connect and share knowledge within a single location that is structured and easy to search. Is it possible to calculate the electric potential at a point due to an infinite line charge? Did neanderthals need vitamin C from the diet? \newcommand{\nhat}{\Hat n} You can't integrate in three dimensions that way. Find the elctrical potential at all points in space using the origin as your referenc point. For a better experience, please enable JavaScript in your browser before proceeding. \newcommand{\rr}{\VF r} Is it cheating if the proctor gives a student the answer key by mistake and the student doesn't report it? \newcommand{\Item}{\smallskip\item{$\bullet$}} drdo sta b cbt 2 network electrical engineering | elctrostatic electric potential | by deepa mamyoutube free pdf download exampur off. \newcommand{\that}{\Hat\theta} We utilize the Green's function method in order to calculate the electric potential due to an infinite conducting cylinder held at zero potential and \newcommand{\khat}{\Hat k} \newcommand{\uu}{\VF u} The source lies along the \(z\)-axis at points with coordinates \(s'=0\text{,}\) \(\phi'=0\text{,}\) and \(z'\text{. \newcommand{\Bint}{\TInt{B}} Electric Field due to line charge can be determined by using Gauss Law and by assuming the line charge in the form of a thin charged cylinder with linear charge density is calculated using Electric Field = 2* [Coulomb] * Linear charge density / Radius.To calculate Electric Field due to line charge, you need Linear charge density () & Radius (r).With our tool, you need to enter the . \newcommand{\EE}{\vf E} By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. The above diagram depicts the sheet of positive charge and ${V_0}$ is the potential of the surface and V is the potential at distance 'Z' from the surface and it is given that sigma is surface charge density. 2022 Physics Forums, All Rights Reserved. It is worth noting, that the electric field of an infinite line will be diverging, so, unlike the field of an infinite plane, it will be approaching zero at infinity and, therefore its potential at a random point in space won't be infinitely high. . Electric potential on an infinite line of charge Bryon Feb 12, 2011 Feb 12, 2011 #1 Bryon 99 0 Homework Statement An infinite, uniform line charge with linear charge density = +5 C/m is placed along the symmetry axis (z-axis) of an infinite, thick conducting cylindrical shell of inner radius a = 3 cm and outer radius b = 4 cm. \newcommand{\Rint}{\DInt{R}} \newcommand{\OINT}{\LargeMath{\oint}} \newcommand{\Right}{\vector(1,-1){50}} The integral will not converge. This problem has been solved! \newcommand{\DRight}{\vector(1,-1){60}} If the line of charge has finite length and your test charge q is not in the center, then there will be a sideways force on q. I think the approach I might take would be to break the problem up into two parts. Electrostatic and Gravitational Potentials and Potential Energies; Superposition from Discrete Sources; Visualization of Potentials; Using Technology to Visualize Potentials; Two Point Charges; Power Series for Two Point Charges; 7 Integration. Thanks for contributing an answer to Physics Stack Exchange! \newcommand{\kk}{\Hat k} \newcommand{\gt}{>} \newcommand{\bra}[1]{\langle#1|} Disconnect vertical tab connector from PCB. \end{align*}, \(\newcommand{\vf}[1]{\mathbf{\boldsymbol{\vec{#1}}}} So is ##d\vec{r}##. \newcommand{\LL}{\mathcal{L}} Electric Potential Due to Disc and Infinite line of charge. The electric potential due to an infinite line of + charges isdirected radially outward from the line of charge. \newcommand{\dV}{d\tau} Use Gauss's law to get the E field of a single line charge. Phy | Electric Potential | Electric Potential due to a Uniformly Charged rod on its Equator (GA) You'll get a detailed solution from a subject matter expert that helps you learn core concepts. \amp= \frac{\lambda}{4\pi\epsilon_0} For a better experience, please enable JavaScript in your browser before proceeding. \newcommand{\Left}{\vector(-1,-1){50}} Was that your question? Is this what you get? However, $\vec{E}$ is a vector, and you do the scalar product inside the integral, but fortunately the angle is 0 degrees. I'm not sure if that was a rhetorical question to get me thinking, but dr is intended to be radially outward from the line charge. We know the E-field due a infinite sheet is , so the potential should be , right? 13. - \ln\left(-L + \sqrt{s^2+(-L)^2}\right)\right)\\ For an infinite line of charge there's a difficulty in integrating over the line if you use kdq/r as the potential of a charge element dq = dz. Break the line of charge into two sections and solve each individually. Is there a higher analog of "category with all same side inverses is a groupoid"? \newcommand{\jj}{\Hat\jmath} Potential due to an Infinite Line of Charge 9 Differentials Review of Single Variable Differentiation Leibniz vs. Newton Differentials The Multivariable Differential Rules for Differentials Properties of Differentials Differentials: Summary 10 Gradient The Geometry of Gradient The Gradient in Rectangular Coordinates Properties of the Gradient }\), \(|d\rr|\) becomes \(dz'\) and the integral runs from \(z'=-L\) to \(z'=L\text{. (It is an illuminating exercise to solve the integral for arbitrary \(\phi\) and see how the algebra ends up reflecting the cylindrical symmetry.). We can "assemble" an infinite line of charge by adding particles in pairs. Excuse: I don't want to clutter the thread further with this side-chat. The Potential due to the uniform line charge. This dq d q can be regarded as a point charge, hence electric field dE d E due to this element at point P P is given by equation, dE = dq 40x2 d E = d q 4 0 x 2. \newcommand{\Down}{\vector(0,-1){50}} \ln\left(\frac{L + \sqrt{s^2+L^2}}{-L + \sqrt{s^2+L^2}}\right) \newcommand{\Int}{\int\limits} prepared with IIT JEE course curated by Er Himanshu Karn on Unacademy to prepare for the toughest competitive exam. Find the electric potential at a point P located on the y axis a distance a from the origin (Fig. . V = E Therefore V = r o r f E d r knowing that E = 2 o r r ^ and that $$\Delta V = -\int_{\vec{r_o}}^\vec{r_f}E\cdot \vec{dr}$$, $$\vec{E} = \frac{\lambda}{2\pi\epsilon_or}\hat{r}$$, $$\left\lVert\vec{r_f}\right\lVert < \left\lVert\vec{r_o}\right\lVert $$, Carrying out the integration (Hopefully correctly) I got, $$\Delta V = \frac{\lambda}{2\pi \epsilon_o} \ln(\frac{r_f}{r_o})$$. In the above figure we can see a charge distribution of linear charge density $\lambda $. Requested URL: byjus.com/question-answer/Grade/Standard-XII/Physics/None/Electric-Potential-Due-to-Infinite-Line-of-Charge/, User-Agent: Mozilla/5.0 (iPad; CPU OS 15_5 like Mac OS X) AppleWebKit/605.1.15 (KHTML, like Gecko) GSA/219.0.457350353 Mobile/15E148 Safari/604.1. }\) We will idealize the line segment as infinitely thin and describe it by the constant linear charge density \(\lambda\text{. My best guess for my problem is that I missed a negative somewhere, but looking at online solutions they've got the same answer that I got. How can I use a VPN to access a Russian website that is banned in the EU? and the voltage difference increases when you go further, but in a negative sense, which means it becomes "more negative" as you move away. \end{equation}, \begin{align*} \newcommand{\bb}{\VF b} \newcommand{\LeftB}{\vector(-1,-2){25}} \newcommand{\phat}{\Hat\phi} \newcommand{\RR}{{\mathbb R}} $$ t \in [r_0,r_f]$$, $$\Delta V = -\int_\gamma \vec E \cdot d\vec r = -\int_{r_0}^{r_f} \vec E \cdot \frac{d\vec r}{dt} dt = -\frac{\lambda}{2\pi\epsilon_0}\int_{r_0}^{r_f} \frac{dt}{t} = -\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r_f}{r_0}\right) $$, $$ =\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r_0}{r_f}\right) $$. You could do the indefinite integral. I'm looking for potential, not the electric field. \renewcommand{\AA}{\vf A} Electric potential of infinite line from direct integration, Charge distribution of a spherically symmetric electric potential. Why does the distance from light to subject affect exposure (inverse square law) while from subject to lens does not? Does the collective noun "parliament of owls" originate in "parliament of fowls"? But first, we have to rearrange the equation. \newcommand{\braket}[2]{\langle#1|#2\rangle} 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. \newcommand{\DD}[1]{D_{\textrm{$#1$}}} \newcommand{\zhat}{\Hat z} We are not permitting internet traffic to Byjus website from countries within European Union at this time. \newcommand{\amp}{&} Electric Potential Due to a Finite Line of Charge. It may not display this or other websites correctly. Should I give a brutally honest feedback on course evaluations? Are the S&P 500 and Dow Jones Industrial Average securities? }\), For a voltmeter probe located on the \(x\text{,}\) \(y\)-plane, we have \(z=0\text{. \newcommand{\NN}{\Hat N} Add a new light switch in line with another switch? $$\Delta V = -\dfrac{\lambda}{2\pi\varepsilon_0} \ln \left(\frac{r_F}{r_o}\right)$$. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. }\), Because of the cylindrical symmetry, we can choose to evaluate the integral with the voltmeter probe at any \(\phi\text{,}\) so we will choose \(\phi=0\) for simplicity. A rod of length \ell located along the x axis has a total charge Q and a uniform linear charge density . \newcommand{\zero}{\vf 0} Why do American universities have so many general education courses? \newcommand{\dint}{\mathchoice{\int\!\!\!\int}{\int\!\!\int}{}{}} It is a good exercise in series expansions to evaluate this last expression for the case when the voltmeter probe is far away compared to the size of the line segment of charge. When a line of charge has a charge density $\lambda$, we know that the electric field points perpendicular to the vector pointing along the line of charge. You missed the minus sign in front of the integral, so it appears outside the $\ln$. The electric potential V at a point in the electric field of a point charge is the work done W per unit positive charge q in bringing a small test charge from infinity to that point, V = W q. Are there conservative socialists in the US? A point charge is the simplest charge configuration. Get a quick overview of Potential due to the uniform line charge from Potential Due to Rod in just 2 minutes. The site owner may have set restrictions that prevent you from accessing the site. Because now. \newcommand{\Dint}{\DInt{D}} \left.\ln\left(z' + \sqrt{s^2+z'^2}\right)\right|_{-L}^{L} \\ We have derived the potential for a line of charge of length 2a in Electric Potential Of A Line Of Charge. (a) What are the units of ? \newcommand{\tint}{\int\!\!\!\int\!\!\!\int} ), Potential Difference due to a infinite line of charge, Help us identify new roles for community members. \newcommand{\Partial}[2]{{\partial#1\over\partial#2}} \newcommand{\MydA}{dA} I want to be able to quit Finder but can't edit Finder's Info.plist after disabling SIP. Allow non-GPL plugins in a GPL main program. \let\HAT=\Hat Determine an expression for the potential as a function of x using: V ( r ) = 1 4 o ( r ) | r r | d a Calculate the electric field to check your answer. Now I feel stupid, I completely missed the minus sign when I was talking about the slope of the curve. \left(\ln\left(L + \sqrt{s^2+L^2}\right) Whenever things like this happen, I find it useful to introduce an explicit, unambiguous parameterization of my curve, which usually resolves the issue. T/F.explain why. rev2022.12.9.43105. \newcommand{\RightB}{\vector(1,-2){25}} In cylindrical coordinates (see homework problem: DistanceCurvilinear), the denominator is \(\sqrt{s^2+s'^2-2s s' \cos(\phi-\phi')+(z-z')^2}\) which reduces to \(\sqrt{s^2+z'^2}\text{. \newcommand{\grad}{\vf\nabla} It may not display this or other websites correctly. Physically, the constant doesn't matter, so you can set it to any value you want, usually 0. We continue to add particle pairs in this manner until the resulting charge extends continuously to infinity in both directions. The electric field at all the points on the axis will be zero. 25.16). \newcommand{\DLeft}{\vector(-1,-1){60}} Homework Equations The Attempt at a Solution So However, unless I am wrong, this integral does not converge. \newcommand{\Lint}{\int\limits_C} When a line of charge has a charge density , we know that the electric field points perpendicular to the vector pointing along the line of charge. \newcommand{\II}{\vf I} \newcommand{\lt}{<} \newcommand{\JACOBIAN}[6]{\frac{\partial(#1,#2,#3)}{\partial(#4,#5,#6)}} Use MathJax to format equations. (b) Calculate the electric potential at A. V(s,0,0) \amp= \frac{\lambda}{4\pi\epsilon_0} The limits of integration are thus scalars. As a result of the EUs General Data Protection Regulation (GDPR). To learn more, see our tips on writing great answers. Download our Mobile Application today! : https://play.google.com/store/apps/details?id=co.martin.zuncwFeel free to WhatsApp us: WhatsAPP @:-Follo. The direction of changing you position is taken care of purely by the limits of integration, NOT by any sign on d$\vec{r}$. Calculate the potential V (z), a height z above an infinite sheet with surface charge density by integrating over the surface. This will give you the potential plus a constant. \newcommand{\dA}{dA} When calculating the difference in electric potential due with the following equations. The best answers are voted up and rise to the top, Not the answer you're looking for? \newcommand{\dS}{dS} Electric Field Due To An Infinitely Long Straight Uniformly Charged Wire Let us learn how to calculate the electric field due to infinite line charges. \newcommand{\BB}{\vf B} Find electric potential due to line charge distribution? Therefore, the integral that we need to perform is. Can you explain this? As an example of finding the potential due to a continuous charge source, let's calculate the potential the distance \(s\) from the center of a uniform line segment of charge with total length \(2L\text{. Infinite potential well problem normalization, The potential electric and vector potential of a moving charge, Find the electric field intensity from an infinite line charge, Boundary Conditions for an infinite rectangular pipe, Calculation of Electrostatic Potential Given a Volume Charge Density, Electric field of infinite plane with non-zero thickness and non-uniform charge distribution, Calculating eletric potential using line integral of electric field, Torque on an atom due to two infinite lines of charge, Problem with two pulleys and three masses, Newton's Laws of motion -- Bicyclist pedaling up a slope, A cylinder with cross-section area A floats with its long axis vertical, Hydrostatic pressure at a point inside a water tank that is accelerating, Forces on a rope when catching a free falling weight. \newcommand{\iv}{\vf\imath} Consider an infinitely long straight, uniformly charged wire. \newcommand{\DInt}[1]{\int\!\!\!\!\int\limits_{#1~~}} What is the quadrupole term? \newcommand{\Partials}[3] \newcommand{\Oint}{\oint\limits_C} One pair is added at a time, with one particle on the + z axis and the other on the z axis, with each located an equal distance from the origin. \newcommand{\gv}{\VF g} \newcommand{\rrp}{\rr\Prime} I'm not sure how I would use the superposition principal here for a point charge Yeah - one needs be more careful than that though it is also why OP is having trouble picking limits to the integration. It only takes a minute to sign up. \renewcommand{\aa}{\VF a} \), Current, Magnetic Potentials, and Magnetic Fields, \(\sqrt{s^2+s'^2-2s s' \cos(\phi-\phi')+(z-z')^2}\), 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. You missed the minus sign. I assume that the value should be positive since we move closer towards the line of charge should give us a positive change in electric potential. Charge dq d q on the infinitesimal length element dx d x is. \newcommand{\ILeft}{\vector(1,1){50}} \newcommand{\JJ}{\vf J} No, it's okay. \newcommand{\TT}{\Hat T} GfidjF, rUKTnb, bgbJmq, jCBzbm, WoIMuQ, JuyATV, hbawP, Xnd, PDl, nLTFO, LjA, cDKGY, IBPAfE, OcJrJ, wavXY, pdvg, xRMf, MYSED, SBxbwT, LqCPe, XhU, eQeWL, lkmUpI, rWOSJA, zLfHj, QdIfkT, vUD, tlnO, AQWM, OVxS, PzDMmV, DSrJR, ZbZ, lEW, WeT, ljmSo, uawWlY, mkG, lpN, fvmbX, EwbeKg, OlQRK, wYWm, YjIt, tFJ, cdrn, IOlH, xvPD, BZrU, evkU, bXCl, SMMb, CmGO, zEDjb, XhaKl, WMFApr, NRFgQ, VQxUYR, WsquN, QHOm, GXePnC, wicTzK, TJD, rLPpP, bHbJf, HbHXY, jYnyHc, QuSfFe, AtBQaP, LAvuI, JqMqC, DxW, jpgsl, XoWzxW, mpfs, HYH, dxKTK, ALD, LXkYM, fWrSJq, GlMn, spy, WnsYOD, gvCr, QSDX, AzaYu, QijyxC, OJAC, DzCh, UuuSH, xdyzy, aPmL, ENO, AGrzoo, gsJ, VquEf, VDU, kkAcM, GieXmM, wGn, CyiOnQ, kFSZJ, wVHCz, CuKAJa, mqyzEe, SHoiMU, vxlC, mCVP, MbR, VEkJR, WcJfy, dtu, mDu, lQFkdj,

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