However, since you are asking for a more formal answer, I will write one. $$ The best answers are voted up and rise to the top, Not the answer you're looking for? The question you must always ask when you use the word "big" is "big with respect to what?" the sun. Just because it. (i) Outside the shell (ii) Inside the shell Easy View solution > Two parallel large thin metal sheets have equal surface charge densities (=26.410 12c/m 2) of opposite signs. The electric field from a thin conducting large plate is Ei = qi / (2Ae_0) in direction outward, from each side of the plate. When two plates are placed next to each other, an electric field is generated. Why does the equation hold better with points closer to the sheet? Maybe one that uses symmetry? Central limit theorem replacing radical n with n. Asking for help, clarification, or responding to other answers. Are defenders behind an arrow slit attackable? This equation holds well for a finite nonconducting sheet as long as we are dealing with points close to the sheet and not too near its edges. Point charges $+3.0\mu C$ and $+7.0\mu C$ are located at the origin and at the point (0.5m, 0) respectively in the x-y plane. Note also that if this were a conductor, then the electric field would be zero inside the material and Derivation 1 gives the correct answer. What happens if you score more than 99 points in volleyball? When discussing the electric field due to a sheet, the size of sheet is compared to our distance from the sheet. An electric field is defined as the electric force per unit charge. To find the electric field, consider a small element on the sheet located at $(x,y)$ of area $dx dy$. So, the value of electric field due to it will be different from the value of electric field for conducting sphere. The best answers are voted up and rise to the top, Not the answer you're looking for? Compute the electric field at a general point $(x,y,z)$ in space-time. Test your Knowledge on Electric field intensity due to a thin uniformly charged infinite plane sheet 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. Electric field due to a large, non-conducting plate and factors of 2 [closed], Help us identify new roles for community members. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. But, here's the important thing. What properties should my fictional HEAT rounds have to punch through heavy armor and ERA? $$ rev2022.12.11.43106. Japanese girlfriend visiting me in Canada - questions at border control? Electromagnetic radiation and black body radiation, What does a light wave look like? to apply Gauss's theorem we require the direction of electric field at P for this purpose we consider two small surface elements S 1 and S 2 the same distance from O as shown in the figure 2.12 the components d . It may not display this or other websites correctly. How many transistors at minimum do you need to build a general-purpose computer? The ratio $\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}$ can be: In the given figure, the particles have charge, \[{{q}_{1}}=-{{q}_{2}}=100nC\text{ }\]\[\text{and }{{q}_{3}}=-{{q}_{4}}=200nC\],and if the distance, \[a=5.0cm\]. since both are in same direction they are added and we get option 'b'as answer. Thus, when we are sitting close to the sheet, the field takes the form you described above. It is then definitely true, that when we are closer to the sheet, in comparison, the sheet has "grown bigger" and therefore can essentially be considered as an infinite sheet and the edge effects can be ignored. Maybe one that uses symmetry? Let us now take the limit of small $z$. Thanks for contributing an answer to Physics Stack Exchange! PSE Advent Calendar 2022 (Day 11): The other side of Christmas. for JEE 2022 is part of JEE preparation. to us, but not w.r.t. In this limit, we find The Earth is big w.r.t. THE BOOK says this: "With twice as much charge now on each inner face, the new surface charge density (s) on each inner face is twice s1. 7 07 : 40. Direction of electric field at points on boundary between two dieletrics. Electric field due to conducting and non-conducting plates. What we really care about is if $z/a$ is small. A charged particle having a charge "q" is placed close to a non conducting plate having charge density "d". Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. How does legislative oversight work in Switzerland when there is technically no "opposition" in parliament? The magnitude of the electric field at $(0,0,z)$ due to this element is then (treating the element as a point charge) A charge in space is carried by an electric field that is linked to the charge. make the sheet very very large. What is the electric field in a parallel plate capacitor? Is there something special in the visible part of electromagnetic spectrum? E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{a^2}{z \sqrt{ 2a^2 + z^2 } } \right] Hebrews 1:3 What is the Relationship Between Jesus and The Word of His Power? Irreducible representations of a product of two groups. E = F/q. Thank you! 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. If the sheet on the left is non conducting and have a uniform charge density 3 (sigma) and the one on the right is conducting and has a uniform charge density (sigma). To learn more, see our tips on writing great answers. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Since electric field is a VECTOR, the NET electric field is: E = E1 + E2 = 2 X s/epsilon0. Note also that if this were a conductor, then the electric field would be zero inside the material and Derivation 1 gives the correct answer. I'm trying to derive the electric field due to a single large, thin, non-conducting plate at a point (see figure). If not then what method would I use to find the electric field in this case. E = r 2 o = 0 = R d ( 2 + r 2) 3 / 2 $$ Why is it that potential difference decreases in thermistor when temperature of circuit is increased? It is given as: E = F/Q Where, E is the electric field F is the force Q is the charge The variations in the magnetic field or the electric charges are the cause of electric fields. This electric field will have in general all 3 - components $(E_x, E_y, E_z)$. Why is the field inside a capacitor not the sum of the field produced by each plate? 1 For a non-conducting sheet, the electric field is given by: E = 2 0 where is the surface charge density. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. Two infinite sheets of charges are placed parallel to each other. For a non-conducting sheet, the electric field is given by: $$E = \frac{\sigma}{2\epsilon_0}$$ If there are any complete answers, please flag them for moderator attention. The magnitude of the electric field at $(0,0,z)$ due to this element is then (treating the element as a point charge) That's exactly right for regions outside the conducting plate. How to get the electric field strength of a plate as approximation of a sphere. As far as you are concerned, the sheet is infinite because you can't see the edges. When discussing the electric field due to a sheet, the size of sheet is compared to our distance from the sheet. The charge of this element is $\sigma dx dy$. Then P is . dE = \frac{1}{4\pi \epsilon_0} \frac{\sigma dx dy}{x^2 + y^2 + z^2 } Use logo of university in a presentation of work done elsewhere. Why does my stock Samsung Galaxy phone/tablet lack some features compared to other Samsung Galaxy models? Electric field of not-grounded conducting plate with a given potential? Work them all out and show that in the small $z/a$ limit, $E_x$ and $E_y$ vanish, while $E_z$ goes to $\frac{\sigma}{2\epsilon_0}$. The only dimensionless number that I can construct using $z$ is $\frac{z}{a}$. Of course, if it were a conductor, then there must be an equal amount of charge on the right surface of the conducting plate. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. to us, but not w.r.t. E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{a^2}{z \sqrt{ 2a^2 + z^2 } } \right] If the plate were a conducting plate (part of a capacitor), would it still be valid to consider the effect of the electric field due to the left side, on any point towards the right in derivation 2 (since electric field does not exist within the volume of the conductor, and therefore cannot propagate through it)? dE_z = \frac{1}{4\pi \epsilon_0} \frac{\sigma z dx dy}{\left( x^2 + y^2 + z^2 \right)^{3/2}} I only to described the simplest possible case to explain my point. Thanks for the answer, @xXx_69_SWAG_69_xXx! You can keep the Gaussian surface inside the material, but there IS an electric field in there, just as you've drawn in the Derivation 2. Inserting a dielectric in a parallel-plate capacitor, MOSFET is getting very hot at high frequency PWM. The distance between the plates in the diagram above is 0.14 meters. Is it possible to hide or delete the new Toolbar in 13.1? The UNIFORM electric field between the leaves would have a magnitude of. Make $a$ large compared to $z$, i.e. As far as you are concerned, the sheet is infinite because you can't see the edges. A similar thing happens here. I had read that thread before posting but was unable to find the exact reason as to why the Gauss Law application in the 1st derivation was incorrect. Why does the equation hold better with points closer to the sheet? Thus, when we are sitting close to the sheet, the field takes the form you described above. 11 : 56. electric field due to thin sheet (non conducting) and conducting plate why it is different. MathJax reference. Making statements based on opinion; back them up with references or personal experience. Counterexamples to differentiation under integral sign, revisited. However, since you are asking for a more formal answer, I will write one. The charge enclosed is the same in both pictures, and the flux is 2EA in both pictures. Expressing the frequency response in a more 'compact' form. However, $z$ is a dimensionfull quantity, and you can't discuss the largeness or smallness of dimensionfull quantities, only dimensionless numbers. As you mention in the question the second derivation is what gives us the correct answer for Electric Field due to this large-thin sheet, and is how its done in most all textbooks. What we really care about is if $z/a$ is small. I understand why the approximation worsens near the edges (because symmetry fails and causes fringe effects) but why is the approximation better near the sheet? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The electric field due to ONE plate is E1 = s/epsilon0. To find the electric field, consider a small element on the sheet located at $(x,y)$ of area $dx dy$. One more thing - your proof calculates the field at $(0,0,z)$ - does this work for other points too? Direction of electric field due to infinite charged sheet: Suppose is the surface charge density on the charge sheet and at point P we have to find the intensity of electric field . E = V/d. We get that the y-component of the electric field due to just this little chunk of our plate, the electric field in the y-component, let's just call that sub 1 because this is just a little small part of the plate. Make $z$ small compared to $a$, i.e. The work done by the field in the above process is: NEET Repeater 2023 - Aakrosh 1 Year Course, To Measure the Thickness of a Given Sheet Using Screw Gauge, Potential Energy of Charges in an Electric Field, Calculating the Value of an Electric Field, Difference Between Electric Field and Magnetic Field, Relation Between Electric Field and Electric Potential, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. But for a non conducting sphere, the charge will get distributed uniformly in the volume of the sphere. My work as a freelance was used in a scientific paper, should I be included as an author? It only takes a minute to sign up. A point charge q moves from point P to a point S along a path PQRS in a uniform electric field E pointing parallel to the x-axis. Just because I'm closer, it doesn't mean the sheet is any bigger. Of course, if it were a conductor, then there must be an equal amount of charge on the right surface of the conducting plate. Why does the equation hold better with points closer to the sheet? For negative charge the . Your proof shows that in the limit, the magnitude of the field approaches the formula I gave. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. This would give E = 0 inside, and $E = \sigma/\epsilon_0$ outside. Does integrating PDOS give total charge of a system? Why don't you do the computation? Connecting three parallel LED strips to the same power supply. Should teachers encourage good students to help weaker ones? Volt per meter (V/m) is the SI unit of the electric field. 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. move in very close to the sheet. Connect and share knowledge within a single location that is structured and easy to search. E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{1}{(z/a)\sqrt{ 2 + (z/a)^2 } } \right] = \frac{\sigma}{2\epsilon_0} + {\cal O}(z/a) Is this an at-all realistic configuration for a DHC-2 Beaver? The equation F = qE determines the force, where F and E are vector variables, and q is a scalar number. the sun. Charges $25 \mathrm{Q}, 9 \mathrm{Q}$ and $\mathrm{Q}$ are placed at point $\mathrm{ABC}$ such that $\mathrm{AB}=4 \mathrm{~m}, \mathrm{BC}=3 \mathrm{~m}$ and angle between $\mathrm{AB}$ and $\mathrm{BC}$ is $90^{\circ} .$ then force on the charge $\mathrm{C}$ is: Why must electrostatic fields at the surface of a charged conductor be normal to the surface at every point? What is the probability that x is less than 5.92? I don't really get the analogy you gave above. Finding the general term of a partial sum series? Why don't you do the computation? I only to described the simplest possible case to explain my point. Gauss's law and superposition for parallel plates. Fair enough. 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If charge A, B, C, D, E and F are \[2\mu C\], \[2\mu C\], \[2\mu C\], \[-2\mu C\], \[-2\mu C\] and \[-2\mu C\] respectively. The first derivation is incorrect because we assume the sheet of charge to be infinitely thin and the surface you are using to apply Gauss Law is also infinitely thin, and so the Gaussian surface must either contain the charged sheet (as it does in derivation 2), or it doesn't contain the second sheet, in which case $Q_{enc}=0$ and so Gauss Law doesn't do anything for us, since we just get $0=0$. Is it appropriate to ignore emails from a student asking obvious questions. mathOgenius. $$ Now, there are two ways to make this small -. This electric field exists even if the plates are not conducting. This electric field will have in general all 3 - components $(E_x, E_y, E_z)$. (3D model). The electric field between parallel plates is influenced by plate density, which determines how large the plate is. Proof that if $ax = 0_v$ either a = 0 or x = 0. We didn't really care if $z$ itself is small (that sentence doesn't even make sense). For a better experience, please enable JavaScript in your browser before proceeding. Why does my stock Samsung Galaxy phone/tablet lack some features compared to other Samsung Galaxy models? Find the electric field at points: Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities ${{\rho }_{1}}$ and ${{\rho }_{2}}$ respectively, touch each other. How do I put three reasons together in a sentence? Examples of frauds discovered because someone tried to mimic a random sequence. So, when I say, $z$ is small, I really mean $\frac{z}{a}$ is small. E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{1}{(z/a)\sqrt{ 2 + (z/a)^2 } } \right] = \frac{\sigma}{2\epsilon_0} + {\cal O}(z/a) Formulas used: Why is the overall charge of an ionic compound zero? In this limit, we find Using Gauss's law derive an expression for the electric field intensity due to a uniform charged thin spherical shell at a point. $$ Correctly formulate Figure caption: refer the reader to the web version of the paper? 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. I have another query. I don't really get the analogy you gave above. I computed the field at $(0,0,z)$ so that I have enough symmetry to say $E_x = E_y = 0$ even for a finite plate. This equation holds well for a finite nonconducting sheet as long as we are dealing with points close to the sheet and not too near its edges. The electric field due to the OTHER is the same: E2 = s/epsilon0. The $z$-component of this electric field is Really nice explanation! Images produced by myself using this website. What are the x components of force? Inside the plate, the field contributions cancel $\vec{E}_{in} = \frac{\sigma}{2\epsilon_0}\hat{x} - \frac{\sigma}{2\epsilon_0}\hat{x} = 0$. Is the electric field at the edge of a uniformly charged disk infinite? Suppose, we wish to find the electric field at a point $(0,0,z)$. The field between plate A and plate B is */*0 if they are charged to some extent, and 0 if they are not. Japanese girlfriend visiting me in Canada - questions at border control? Thank you! Just because it. 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 . $$ I computed the field at $(0,0,z)$ so that I have enough symmetry to say $E_x = E_y = 0$ even for a finite plate. At a different point, there is no symmetry, so $E_x , E_y \neq 0$ which only makes the computation more complicated. The Question and answers have been prepared according to the JEE exam syllabus. Connect and share knowledge within a single location that is structured and easy to search. Find the force of attraction between them? How can I fix it? If you are close to the sheet, the edge effects are negligible. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centre of the spheres is zero. An electric field is an area or region where every point of it experiences an electric force. Compute the electric field at a general point $(x,y,z)$ in space-time. For a given closed surface . The electrons are attracted to the plate with the opposite charge. dE_z = \frac{1}{4\pi \epsilon_0} \frac{\sigma z dx dy}{\left( x^2 + y^2 + z^2 \right)^{3/2}} This equation holds well for a finite nonconducting sheet as long as we are dealing with points close to the sheet and not too near its edges. Why is the y-component of electric field of a uniformly-charged disk near its center the same as that of infinite sheet of charge? What is the electric field in a parallel plate capacitor? The coordinates of P, Q, R and S are (a,b,0), (2a,0,0), (a,-b,0) and (0,0,0). So if it were a conducting plate, can we say that each side of the plate produces an electric field E = /20, and that the net E at any point will be equal Enet = /20 + /20 (since both sides produce an outward electric field?). If the area on both plates is 1m^2 then calculate the value of electric field at (a) to the . Just because I'm closer, it doesn't mean the sheet is any bigger. This creates a force between the plates. where $\sigma$ is the surface charge density. Please help the asker edit the question so that it asks about the underlying physics concepts instead of specific computations. So I would say that your mistake is that you did NOT draw the electric field going to the right inside the material in your first figure (Derivation 1). I've referred some textbooks, and they say that the result of the 2nd derivation is correct. By symmetry, this electric field will point solely in the $z$-direction. How many transistors at minimum do you need to build a general-purpose computer? Make $a$ large compared to $z$, i.e. The charge of this element is $\sigma dx dy$. We didn't really care if $z$ itself is small (that sentence doesn't even make sense). Make $z$ small compared to $a$, i.e. The $z$-component of this electric field is Since it's a nonconducting plate, the charge sits only on the left surface and there is indeed an electric field inside the material (we're ignoring dielectric effects here, right? Now, there are two ways to make this small -. $$ It is then definitely true, that when we are closer to the sheet, in comparison, the sheet has "grown bigger" and therefore can essentially be considered as an infinite sheet and the edge effects can be ignored. $$ But, here's the important thing. dE = \frac{1}{4\pi \epsilon_0} \frac{\sigma dx dy}{x^2 + y^2 + z^2 } Suppose, we wish to find the electric field at a point $(0,0,z)$. CGAC2022 Day 10: Help Santa sort presents! Also It would be greate if anyone can comment on how to find the electric field by directly solving the poisson equation. At a different point, there is no symmetry, so $E_x , E_y \neq 0$ which only makes the computation more complicated. I would like to know which method is correct, and why is the other method wrong? $$ How is the merkle root verified if the mempools may be different? @Prahar Could you please give me a more formal explanation? Using both equations, we can determine the electric sheet due to the charged sheet which will also give us the relation between electric field and distance from the sheet. Can a prospective pilot be negated their certification because of too big/small hands? Better way to check if an element only exists in one array. Integrating this over the sheet, we find the total electric field at $(0,0,z)$ as Prove that isomorphic graphs have the same chromatic number and the same chromatic polynomial. It only takes a minute to sign up. What is this fallacy: Perfection is impossible, therefore imperfection should be overlooked. Consider a square sheet with edges located at $(a,0)$, $(-a,0)$, $(0,a)$ and $(0,-a)$. Definition of Gaussian Surface Imagine sitting very close to the sheet. Can I change any equation/assumption in the wrong method to arrive at the right result? Integrating this over the sheet, we find the total electric field at $(0,0,z)$ as Let us now take the limit of small $z$. Both the statements above are completely equivalent. For a non-conducting sheet, the electric field is given by: $$E = \frac{\sigma}{2\epsilon_0}$$ Can we keep alcoholic beverages indefinitely? How can I use a VPN to access a Russian website that is banned in the EU? The question you must always ask when you use the word "big" is "big with respect to what?" Electric field due to negatively charged plate towards that plate and is equal to sigma/ 2ephslanot.electic field due to positively charged plate is away from it and is equal to Sigma/2 ephslano. You are using an out of date browser. Connecting three parallel LED strips to the same power supply. JavaScript is disabled. Why doesn't the method of images work for this problem? Help us identify new roles for community members. Work them all out and show that in the small $z/a$ limit, $E_x$ and $E_y$ vanish, while $E_z$ goes to $\frac{\sigma}{2\epsilon_0}$. Homework Statement. Your proof shows that in the limit, the magnitude of the field approaches the formula I gave. Integration of the electric field then gives the capacitance of conducting plates with the corresponding geometry. Electric field due to conducting and non-conducting plates, A very large nonconducting plate lying in the xy-plane carries a charge, electric field due to thin sheet (non conducting) and conducting plate why it is different, Electric Field Due To Infinite Plane Sheets(Conduction and Non Conducting) -Derivation, Electric field due to conducting and non-conducting sheet | JEE & NEET. Imagine sitting very close to the sheet. Electric Field Intensity Due to Non-Conducting Sphere The charge on the conducting sphere get distributed over the surface. How does the Chameleon's Arcane/Divine focus interact with magic item crafting? Give reason. I have spent HOURS on the internet but the sites I have found do not clearly distinguish between PLATES and CONDUCTING PLATES. Is the EU Border Guard Agency able to tell Russian passports issued in Ukraine or Georgia from the legitimate ones? Hence, the flux is the integration of electric field vectors and area vectors. Electric field lines fall within a circle? rev2022.12.11.43106. Inconsistent image charges: what happens when three conducting planes meet? $$ And the voltage between the plates is 28 volts. Is it correct to say "The glue on the back of the sticker is dying down so I can not stick the sticker to the wall"? 4 . Consequently if we take case of finite disk the following is the resulting integration. Thus, the electric field is any physical quantity that takes different values of electric force at different points in a given space. Consider a square sheet with edges located at $(a,0)$, $(-a,0)$, $(0,a)$ and $(0,-a)$. otherwise you'll need to know the dielectric constant of the material.) Z13 Physics Y Kumar Dehradun. move in very close to the sheet. The electric field is created by the movement of electrons within the plates. However, how do we know the. Could an oscillator at a high enough frequency produce light instead of radio waves? $$ $$ At what point in the prequels is it revealed that Palpatine is Darth Sidious? 428 . Use MathJax to format equations. Electric Field Due To Infinite Plane Sheets(Conduction and Non Conducting) -Derivation. in this video, we will study about electric field due to #conducting_and_nonconducting_sheet *all doubts explained success router | physics by sanjeet singh | sanjeet singh iit (ism). However, $z$ is a dimensionfull quantity, and you can't discuss the largeness or smallness of dimensionfull quantities, only dimensionless numbers. Classic electrostatics image problem surface charge. By the gauss law, flux is charge divided by absolute permittivity. The Earth is big w.r.t. At what point in the prequels is it revealed that Palpatine is Darth Sidious? One more thing - your proof calculates the field at $(0,0,z)$ - does this work for other points too? The only dimensionless number that I can construct using $z$ is $\frac{z}{a}$. I'm solving it using 2 methods, and arriving at a different answer using both. It is equal to the electric field generally, the magnitude of the electric field from this point, times cosine of theta, which . However, how do we know the. Electrical Force And Its Characteristics 15,399 Stay tuned with BYJU'S for more such interesting derivations in physics, chemistry and maths in an engaging way with video explanations. We assume positive charge in the formulas. How could my characters be tricked into thinking they are on Mars? What is the formula for an electric field? If you are close to the sheet, the edge effects are negligible. As an alternative to Coulomb's law, Gauss' law can be used to determine the electric field of charge distributions with symmetry. @Prahar Could you please give me a more formal explanation? Electric field due to plate = d/2epsilon hence force = Eq = dq/2epsilon . Thanks for the reply @Qmechanic. A similar thing happens here. Tabularray table when is wraped by a tcolorbox spreads inside right margin overrides page borders. LET'S LEARN PHYSICS. make the sheet very very large. $$ $$ But anyways I managed to solve it. The electric field between two plates: The electric field is an electric property that is linked with any charge in space. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Both the statements above are completely equivalent. So, when I say, $z$ is small, I really mean $\frac{z}{a}$ is small. $$ How is dielectric constant both $E_{net}/E_o$ and $/_o$? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Fair enough. where $\sigma$ is the surface charge density. I understand why the approximation worsens near the edges (because symmetry fails and causes fringe effects) but why is the approximation better near the sheet? Why is this integral for a uniform electric field of a charged plate not evaluating correctly? $$ electric field due to non conducting plate / sheet (in English ) 78 views Jan 1, 2021 this video drives an expression for electric field due to infinite long uniformly charged. This would give E = 0 inside, and E = / 0 outside Share Cite By symmetry, this electric field will point solely in the $z$-direction. 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