# electric field lines formula

Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. What is the earliest time we might be reasonably sure who won the US 2020 Presidential election? , \tag{e4}\label{e4} Since it is a finite line segment, from far away, it should look like a point charge.

\frac{d}{dx}\left(\frac{x + a}{r_1}\right) - \frac{d}{dx}\left(\frac{x + a}{r_1}\right) = 0, However, the actual electric field in that region exists at every point in space.

Inthissection,wepresentanotherapplication–theelectricfieldduetoaninfinitelineofcharge.5.6:CalculatingElectricFieldsofChargeDistributions,["article:topic","authorname:openstax","ContinuousChargeDistribution","infiniteplane","infinitestraightwire","linearchargedensity","surfacechargedensity","volumechargedensity","license:ccby","showtoc:no"],CreativeCommonsAttributionLicense(by4.0),Explainwhatacontinuoussourcechargedistributionisandhowitisrelatedtotheconceptofquantizationofcharge,Describelinecharges,surfacecharges,andvolumecharges,Calculatethefieldofacontinuoussourcechargedistributionofeithersign.Thisfieldishowonechargeexertsaforceonanotheroveradistance.Wewillnowsolveequation\eqref{e2}.thesecondfactortogetFormulaofElectricField.\nonumber\],\begin{align*}\vec{E}(P)&=\dfrac{1}{4\pi\epsilon_0}\int_0^{L/2}\dfrac{2\lambdadx}{(z^2+x^2)}\dfrac{z}{(z^2+x^2)^{1/2}}\hat{k}\\[4pt]&=\dfrac{1}{4\pi\epsilon_0}\int_0^{L/2}\dfrac{2\lambdaz}{(z^2+x^2)^{3/2}}dx\hat{k}\\[4pt]&=\dfrac{2\lambdaz}{4\pi\epsilon_0}\left[\dfrac{x}{z^2\sqrt{z^2+x^2}}\right]_0^{L/2}\hat{k}.Wewanttohearfromyou.However,don’tconfusethiswiththemeaningof$$\hat{r}$$;weareusingitandthevectornotation$$\vec{E}$$towritethreeintegralsatonce.Thefirstorderofbusinessistoconstraintheformof$${\bfD}$$usingasymmetryargument,asfollows.\nonumber,Tosolvesurfacechargeproblems,webreakthesurfaceintosymmetricaldifferential“stripes”thatmatchtheshapeofthesurface;here,we’lluserings,asshowninthefigure.Wecontinuetoaddparticlepairsinthismanneruntiltheresultingchargeextendscontinuouslytoinfinityinbothdirections.Thisisaverycommonstrategyforcalculatingelectricfields.\frac{1}{r_1}-\frac{x+a}{r_1^2}\frac{\partialr_1}{\partialx}-\frac{1}{r_2}+\frac{x-a}{r_2}\frac{\partialr_2}{\partialx}=0,@BillN$\theta$and$\phi$aremeasuredwithrespecttopositivedirectionofthex-axis.Wewillchecktheexpressionwegettoseeifitmeetsthisexpectation.UseMathJaxtoformatequations.fortheelectricfield.Fieldlinesforthreegroupsofdiscretechargesareshownin.,.Theresultservesasauseful“buildingblock”inanumberofotherproblems,includingdeterminationofthecapacitanceofcoaxialcable(Section5.24).