Electric Field Due To Infinite Plane Sheets (Conduction and Non Conducting) -Derivation - YouTube 0:00 / 7:40 #mathOgenius Electric Field Due To Infinite Plane Sheets (Conduction. 2022 Physics Forums, All Rights Reserved, 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. The field pattern is shown in the adjacent figure. E = \frac{\sigma}{2\epsilon_0} 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. Zero potential is selected in the centre of the plate. For a second-order equation, you need to give two boundary conditions. b) Also determine the electric potential at a distance z from the centre of the plate. -- For x = 6 cm, I only used the electric field of the slab, since I thought it would block the field of the sheet. \[E_p(z)\,=\, - \int^z_{0} \vec{F} \cdot \mathrm{d}\vec{z}\], \[\varphi\,=\, - \int^z_{0} \frac{\vec{F}} {Q}\cdot \mathrm{d}\vec{z}\], \[\varphi\,=\, - \int^z_{0} \vec{E}\cdot \mathrm{d}\vec{z}\], \[\oint_S \vec{E} \cdot \mathrm{d}\vec{S}\,=\, \frac{Q}{\varepsilon_0}\], \[\oint_S \vec{E} \cdot \vec{n}\mathrm{d}S\,=\, \frac{Q}{\varepsilon_0}\], \[2\oint_{S_b} \vec{E} \cdot \vec{n}\mathrm{d}S\,+\,\oint_{S_la} \vec{E} \cdot \vec{n}\mathrm{d}S\,=\, \frac{Q}{\varepsilon_0}\tag{*}\], \[\oint_{S_la} \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,0\], \[ \vec{E} \cdot \vec{n} \,=\, En\,=\,E\], \[\oint_{S_b} \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,\oint_{S_b} E n\mathrm{d}S\,=\, \oint_{S_b} E\mathrm{d}S\], \[\oint_{S_b} \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,E \oint_{S_b} \mathrm{d}S\,=\,ES_p\,,\], \[2 E S_b\,=\, \frac{\varrho a S_b }{\varepsilon_0}\], \[2 E \,=\, \frac{\varrho a}{\varepsilon_0}\], \[E \,=\, \frac{\varrho a}{2 \varepsilon_0}\], \[\oint_S \vec{E} \cdot \vec{n}\mathrm{d}S\,=\, \frac{Q_1}{\varepsilon_0}\,,\], \[\oint_{S_b} \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,\oint_{S_b} E n\mathrm{d}S\,=\, \oint_{S_b} E\mathrm{d}S\,=\,E \oint_{S_b} \mathrm{d}S\], \[\oint_{S_b} \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,E S_b\], \[2 E S_b\,=\, \frac{Q_1}{\varepsilon_0}\], \[2 E S_b\,=\, \frac{ 2z \varrho S_b }{\varepsilon_0}\], \[E \,=\, \frac{\varrho}{\varepsilon_0} \,z\], \[\varphi (z)\,=\, - \int_{0}^z \vec{E} \cdot \mathrm{d}\vec{z}\], \[ \varphi (z)\,=\, - \int^{z}_{0} E \mathrm{d}z \], \[ \varphi (z)\,=\, - \int^{z}_{0}\frac{\varrho}{\varepsilon_0}\,z \mathrm{d}z \], \[ \varphi (z)\,=\, - \frac{\varrho}{\varepsilon_0}\int^{z}_{0}z \,\mathrm{d}z \,=\, -\,\frac{\varrho}{\varepsilon_0} \left[ \frac{z^2}{2}\right]^{z}_{0} \,=\, -\,\frac{\varrho}{\varepsilon_0}\, \frac{z^2}{2}\,.\], \[ \varphi (z)\,=\, -\,\frac{\varrho}{\varepsilon_0}\, \frac{z^2}{2}\,.\], \[\varphi (z)\,=\, - \int_{0}^z \vec{E} \cdot \mathrm{d}\vec{z}\,=\, - \int_{0}^z E \mathrm{d}z\,.\], \[\varphi (z)\,=\, - \int^{\frac{a}{2}}_{0} E_{in}\,\mathrm{d}z \, -\, \int^{z}_{\frac{a}{2}} E_{out}\,\mathrm{d}z\], \[\varphi (z)\,=\, - \int^{\frac{a}{2}}_{0} \frac{\varrho}{\varepsilon_0}\,z \,\mathrm{d}z -\, \int^{z}_{\frac{a}{2}}\frac{\varrho a}{2 \varepsilon_0} \,\mathrm{d}z \,\], \[\varphi (z)\,=\, - \frac{\varrho}{\varepsilon_0}\int^{\frac{a}{2}}_{0}z \mathrm{d}z\, -\, \frac{\varrho a}{2 \varepsilon_0}\int^{z}_{\frac{a}{2}} \mathrm{d}z \,=\,-\,\frac{\varrho}{\varepsilon_0} \left[ \frac{z^2}{2}\right]^{\frac{a}{2}}_{0}\, - \,\frac{\varrho a}{2 \varepsilon_0}\left[z\right]^{z}_{\frac{a}{2}}\], \[\varphi (z)\,=\, - \,\frac{\varrho a^2}{8 \varepsilon_0} \, -\,\frac{\varrho}{2 \varepsilon_0}\, z^2\, +\,\frac{\varrho a^2}{4\varepsilon_0}\], \[\varphi (z)\,= \,\frac{\varrho a^2}{8 \varepsilon_0} \, -\,\frac{\varrho}{2 \varepsilon_0}\, z^2\, \], \[E \,=\, \frac{\varrho a}{2 \varepsilon_0}\,.\], \[E \,=\, \frac{\varrho}{\varepsilon_0} \,z\,.\], \[\varphi (z)\,=\,\frac{\varrho a^2}{8 \varepsilon_0} \, -\frac{\varrho a}{2 \varepsilon_0}\,|z| \,.\], \[\varphi (z)\,=\, -\,\frac{\varrho}{2 \varepsilon_0}\, z^2\,. Dividing both sides by the cross-sectional area A, we can eliminate A on both sides and solving for the electric field, the magnitude of electric field generated by this infinite plate or sheet of charge, E becomes equal to over 2 0. Potential at a distance z from the centre of the plate at point A is equal to negative taken integral of intensity from a point of zero potential to point A. The electric field is uniform outside the plate with intensity \(E\,=\, \frac{\varrho a}{2 \varepsilon_0}\). E = \frac{\sigma}{\epsilon_0} Alternatively, you can reason as follows. Then, field outside the cylinder will be. We choose the Gaussian surface to be a surface of a cylinder with length 2z, its axis being perpendicular to the plate and the centre of the plate passes through the centre of the cylinder. If you wish to filter only according to some rankings or tags, leave the other groups empty. With the exception of points on charged surfaces, the first derivative of the potential is also continuous, i.e. It is known that a non-conducting sheet of charge (charge on only one "side") has electric field magnitude where is the surface charge density in Coulombs per square meter. The conducting slab has a net charge per unit area of 2 = 5 C/m2. The electric field above a uniformly charged non-conducting sheet is E. If the non-conducting sheet is now replaced by a conducting sheet, with the charge same as before, the new electric field at the same point is: A 2E B E C E 2 D None Solution The correct option is B. if we include the interior the symmetry is failed because one side there is electric field other side there is no field itself. In the case of nonconducting sheet, there is no such limitation. Formulas used: We know that the electric field is directed radially outward for a positive charge, and for a negative point charge, the electric field is directed inwards. The relations describing the intensity outside and inside the plate differ. The vector of electric field intensity \(\vec{E}\) is parallel to the \(\vec{z}\) vector. The potential inside the plate is represented by the formula. $\therefore$ Electric field due to an infinite conducting sheet of the same surface density of charge is $ \dfrac{E}{2}$. And I put my Gaussian pill box around the entire sheet. 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, Difference between $E$ field configuration, sheet of charge: infinite sheet of charge, conducting vs. non-conducting, Help us identify new roles for community members, Field between the plates of a parallel plate capacitor using Gauss's Law, Classic electrostatics image problem surface charge, Boundary condition of charge sheet in an external electric field, I don't understand equation for electric field of infinite charged sheet, How to calculate the electric field outside an infinitely long conducting cylinder with surface charge density , Using square loops to calculate electric field of infinite plane of charge, Electric field between oppositely charged metal plates, Concentration bounds for martingales with adaptive Gaussian steps, QGIS expression not working in categorized symbology, Counterexamples to differentiation under integral sign, revisited. for z>a/2), we proceed similarly as in the previous section. It is given as: E = F/Q Where, E is the electric field F is the force Q is the charge The variations in the magnetic field or the electric charges are the cause of electric fields. However, the total electric field near a surface is due to all charges, not just the surface charge you are near to. Ad blocker detected Knowledge is free, but servers are not. calculate the electric field intensity. Gauss's law is easily applicable (I.e. It may not display this or other websites correctly. There is a nice extended explanation including pictures at this site. b) Also determine the electric potential at a distance z from the centre of the plate. This is why the electric field of a non-conducting sheet of charge is half of that of a conducting sheet of charge. The electric intensity increases linearly inside the plate from the centre to the surface of the plate. We examine the field outside and inside the plate separately. Since it is a conducting plate so the charge will be distributed uniformly on the surface of the plate. If he had met some scary fish, he would immediately return to the surface, MOSFET is getting very hot at high frequency PWM. the potential is a smooth function. When calculating the potential outside the plate (i.e. Now, in case of a conductor, you can show that the total electric field is twice this value using Gauss' law. But outside the bulk at any point we can apply superposition . We will let the charge per unit area equal sigma . \end{equation}. Note: The electric potential is always continuous, because it is actually work done by transferring a unit charge and it can not be changed "by steps". E = 2 0 is the electric field due to the surface charge. $E = \frac{\sigma}{2\epsilon_0}$ is the electric field due to the surface charge. The normal component changes "by steps" proportional to the surface charge density. You are using an out of date browser. If you recall that for an insulating infinite sheet of charge, we have found the electric field as over 2 0 because in the insulators, charge is distributed throughout the volume to the both sides of the surface, whereas in the case of conductors, the charge will be along one side of the surface only. of thickness (t) . For a better experience, please enable JavaScript in your browser before proceeding. Due to this symmetry we can also solve the whole task only for positive z values, the only thing that changes for negative z is the direction of the vector of electric intensity. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The outside field is often written in terms of charge per unit length of the cylindrical charge. In order to apply Gauss's law, we first need to draw the electric field lines due to a continuous distribution of charge, in this case a thin flat sheet. You are using an out of date browser. Find the electric field on the axis at But you asked for a "easy to remember" explanation. Why does the USA not have a constitutional court? Electric field and potential near the sheet are Electric field and potential near the sheet are The electric field of a point charge surrounded by a thick spherical shell. The total electric flux is obtained by adding the flux through the lateral area and through both bases of the cylinder. For a better experience, please enable JavaScript in your browser before proceeding. the closed surface integral easily soved) only when there is symmetry in the problem. The intensity in the centre of the plate is zero. We express the charge by using this volume and the charge density Q=V=aSp. for z>a/2. in the centre of the plate). You can check out similar questions with solutions below. Hence, the correct answer is option (B). the charged volume "reaches" to infinity. A solid nonconducting sphere of radius R has a uniform charge distribution of volume charge density, = 0Rr, where 0 is a constant and r is the distance from the centre of the sphere. But in the case of a non-conducting sheet, you just have $\sigma$. In order to obtain the constants I used three things: 1) the fact that the electric field outside the plate is symmetrical w.r.t the plate (and not just constant) 2) Gauss law where the two bases of the Gaussian cylinder/box are outside the plate 3) Gauss law where one base is inside the plate and the other base outside it. The field between the plates is zero. \], \[\mathrm{\Delta} E\,=\, \frac{\mathrm{\Delta} \sigma}{2 \epsilon_0}\,=\, \frac{\varrho \mathrm{\Delta} r}{2 \epsilon_0}\,.\], \[E\,=\, \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\varrho }{2 \epsilon_0}\,\mathrm{d} r\], \[E\,=\,\frac{\varrho }{2 \epsilon_0} \int^{\frac{a}{2}}_{-\frac{a}{2}}\mathrm{d} r\,=\, \frac{\varrho }{2 \epsilon_0}[r]^{\frac{a}{2}}_{-\frac{a}{2}}\,=\, \frac{\varrho }{2 \epsilon_0}\,\left(\frac{a}{2}+\frac{a}{2}\right)\], \[E\,=\, \frac{\varrho a }{2 \epsilon_0}\], \[E\,=\, \int^{z}_{-\frac{a}{2}} \frac{\varrho }{2 \epsilon_0}\,\mathrm{d} r\,-\,\int^{\frac{a}{2}}_{z} \frac{\varrho }{2 \epsilon_0}\,\mathrm{d} r\], \[E\,=\, \frac{\varrho }{2 \epsilon_0}\int^{z}_{-\frac{a}{2}}\mathrm{d} r\,-\, \frac{\varrho }{2 \epsilon_0}\int^{\frac{a}{2}}_{z}\mathrm{d} r \,=\, \frac{\varrho }{2 \epsilon_0}[z]^{z}_{-\frac{a}{2}}\,-\, \frac{\varrho }{2 \epsilon_0}[z]^{\frac{a}{2}}_{z}\], \[E\,=\, \frac{\varrho }{2 \epsilon_0}(z\,+\, \frac{a}{2}\,-\, \frac{a}{2}\,+\,z)\,=\, \frac{\varrho }{2 \epsilon_0}\,2z\], \[E\,=\, \frac{\varrho }{\epsilon_0}\,z\,,\], Tasks requiring comparison and contradistinction, Tasks requiring categorization and classification, Tasks to identify relationships between facts, Tasks requiring abstraction and generalization, Tasks requiring interpretation,explanation or justification, Tasks aiming at proving, and verification, Tasks requiring evaluation and assessment, Two balls on a thread immersed in benzene, Electric Intensity at a Vertex of a Triangle, A charged droplet between two charged plates, Capaciter partially filled with dielectric, Electrical Pendulum in Charged Spheres Field (Big Deflection), Gravitational and electric force acting on particles, Field of Charged Plane Solved in Many Ways, Electric resistance of a constantan and a copper wire, Electrical Resistances of Conductors of Different Lengths, Electrical Resistance of Wires of Different Cross Sections, Measuring of the electrical conductivity of sea water, Two-wire Cable between Electrical Wiring and Appliance, Using Kirchhoffs laws to solve circiut with two power supplies, Change of the current through potentiometer, Application of Kirchhoffs laws for calculation of total resistance in a circuit, Current-carrying wire in a magnetic field, Magnetic Force between Two Wires Carrying Current, Magnetic Field of a Straight Conductor Carrying a Current, Magnetic Field of a Straight Conductor inside a Solenoid, The motion of a charged particle in homogeneous perpendicular electric and magnetic fields, Voltage Induced in a Rotating Circular Loop, Inductance of a Coil Rotating in a Magnetic Field, The Length of the Discharge of the Neon Lamp, Instantaneous Voltage and Current Values in a Series RLC Circuit, RLC Circuit with Adjustable Capacitance of Capacitor, Heating Power of Alternating Current in Resistor, Resonance Frequency of Combined Series-Parallel Circuit. Here, we can see that the electric field has no relation with the distance "r". In this section we determine the intensity of the electric field outside the plate, i.e. Also I believe the questioner intends an infinite nonconducting charged plane and a charged conductor of sufficiently large . We obtain a relation: where Sb is the surface area of the base of Gaussian cylinder. The sum of the two must vanish inside the conductor, therefore just outside the conductor, the field must be double that due to only the local surface charge. The surface is chosen this way, because the vector of electric intensity is perpendicular to the cylinder bases and parallel to the lateral area of the cylinder, which simplifies the calculation of the scalar product. The vectors of electric intensity do not have the same direction. We factor the constants out of the integrals and calculate the integrals. However, the total electric field near a surface is due to all charges, not just the surface charge you are near to. JavaScript is disabled. This is a very easy question, but I often confused myself. Due to symmetrical charge distribution, the easiest way to find the intensity of electric field is using Gauss's law. Continue on app (Hindi) Electrostatics: Coulomb's Law - IIT JEE. For negative values of z the function is negative. It may not display this or other websites correctly. Electric Field: Parallel Plates. for za/2 from the centre of the thick plate, we add contributions of all these thin plates. Since the two cylinder bases at the same distance from the charged plate, the electric intensity vector is on both bases of the same magnitude. Both the electric field dE due to a charge element dq and to another element with the same charge but located at the opposite side of the ring is represented in the following figure. The procedure is similar to that in the previous section, so it is not commented in detail. Mathematica cannot find square roots of some matrices? Multiplying 0 0 by R2 R 2 will give charge per unit length of the cylinder.