The answer is that themagnitudes of the charge of the electron and proton are equal. The magnitude of the electric field is adjusted until thegravitational force Fg = mg = mg j on the oil drop is exactly balanced by the electricFe = qE. It proves to be something called an elliptic integral. This seems to be poor writing on the question on the part of your instructor. Step 1 - Enter the Charge. Electricity and Magnetism lectures series for BS Physics as per HEC Syllabus This lecture explains the electric filed due to continuous charge distribution i. We want to find the total charge of the disc. MathJax reference. Electric Field Due to a Charged Disk, Infinite Sheet of Charge, Parallel Plates - Physics Problems. Your interpretation of this statement is reasonable and it is the only thing that one would expect from such a statement: a plane slab with a cylindrical cut-out, or more specifically charge filling the set The superposition of these two will give the relevant geometry: slab with a charge free cavity. 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. To find dQ, we will need dA d A. is carved out from the slab. 112 17 : 24 . The Organic Chemistry Tutor. The Electric Field Due to a Charged Disk Figure 3a shows a circular disk that is uniformly charged. Electric field due to a uniformly charged disc. (* This is a comment *) and 2. Electric field due to uniformly charged disk. Suppose this balancing occurs whenE = E y j = (1.92 105 N C) j , with E y = 1.92 105 N C .force,theelectricfieldis(a) What is the mass of the oil drop? It seems it's a horrendous job to calculate the the $E$ field at an arbitrary point due to the circular/spherical cavity. The central z axis is perpendicular to the disk face, with the origin at the disk. Stack Exchange Network. (6) Complete the integration to obtain E . Arbitrary shape cut into triangles and packed into rectangle of the same area, PSE Advent Calendar 2022 (Day 11): The other side of Christmas, If he had met some scary fish, he would immediately return to the surface. For example, related to the problem of a a unifiromly charged disk, Purcell and Morin's textbool Electricity and Magnetism reads: It is not quite so easy to derive the potential for general points away The electric dipolemoment vector p points from the negative charge to the positive charge, and has amagnitudep = 2aqThe torque acting on an electric dipole places in a uniform electric field E is = pEThe potential energy of an electric dipole in a uniform external electric field E isU = p EThe electric field at a point in space due to a continuous charge element dq isdE =1dqr4 0 r 2At sufficiently far away from a continuous charge distribution of finite extent, theelectric field approaches the point-charge limit.252.12 Problem-Solving StrategiesIn this chapter, we have discussed how electric field can be calculated for both thediscrete and continuous charge distributions. Calculate the electric field on the axis of the disk at (a) 5.0 cm, (b) 10.0 cm, (c) 50 cm and (d) 200 cm from the center of the. Find the Electric Field due to this charge distribution on the axis of symmetry (z axis) for both z > 0 and z < 0. = Q R2 = Q R 2. Monopole and Dipole Terms of Electric potential (V) on Half Disk. But isn't having to calculate the electric field at any point in space, which in this case would be a suitable superposition of the previous two cases, a bit too much. Just to set this in more permanent form: yes, the task as you have (reasonably) construed it is pretty hopeless. Welcome. Edit: if you try to do the calculations for x < 0 you'll end up in trouble. 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. The magnitude of the charge of the electron and proton ise = 1.6 1019 C . Thus, the ratio of the magnitudes of the electric and gravitationalforce is given by28 1 e2 1 2e2 4 0 r 4 0(9.0 109 N m 2 / C2 )(1.6 1019 C) 2==== 2.2 103922112731 m p me Gm p me (6.67 10 N m / kg )(1.7 10 kg)(9.1 10 kg)G 2 r which is independent of r, the distance between the proton and the electron. The particle has an initial velocity v 0 = v0 iperpendicular to E .Figure 2.13.1 Charge moving perpendicular to an electric field30(a) While between the plates, what is the force on the electron? I'd like to work it out on my own. rev2022.12.11.43106. The total charge of the disk is q, and its surface charge density is (we will assume it is constant). R is greater than 2R. Thanks for contributing an answer to Physics Stack Exchange! Class 12 Physics | Electrostatics | #39 Electric Field due to a Uniformly Surface Charged Disc. If the electric field is known, then the electrostatic force on any charge q q size 12{q} {} is simply obtained by multiplying charge times electric field, or F = q E F = q E size 12{F=qE} {}. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, . Electric Field at an Arbitrary Point due to a Uniformly Charged Disk, physics.stackexchange.com/questions/284147/, Help us identify new roles for community members, Gauss's law for cylinder with infinite height with a spherical cavity. Question: The Electric Field Due to a Charged Disk to answer this question. 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. I think that the easiest way would be to fill in the cavity and calculate the field at a point. We will calculate the electric field due to the thin disk of radius R represented in the next figure. Electric Field Due to charged disk kdqz (2 +x ) /2 Using the equation for a ring. Using our force laws, we have0 = mg + qE mg = qE yWith the electrical field pointing downward, we conclude that the charge on the oil dropmust be negative. (a) Show that the electric field of the dipole in the limit where ra is32Ex =3ppsin cos , E y =3cos 2 1)33 (4 0 r4 0 rwhere sin = x / r and cos = y / r . 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The best experiments showthat the difference between these magnitudes is a number on the order of 10 24 . z = 3 x E Gdq dr FK $$S_3=\{(x,y,z)\in\mathbb R^3: -dR^2\},$$ The actual formula for the electric field should be. 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. Design by 123DOC, Xem v ti ngay bn y ca ti liu ti y (29.25 MB, 2,361 trang ), Study materials for MIT course 8 02t electricity and magnetism FANTASTIC MTLS, NHNG YU T MI TRNG TC NG N VIC M NG BAY SI GN THNG HI. Just like here we assumed the disc to be made up of many infinitesimally thin discs, we can use the same iea to calculate the electric field at a point due to a charged hollow cylinder. (i) Equipotential surfaces due to single point charge are concentric sphere having charge at the centre. Examples of frauds discovered because someone tried to mimic a random sequence. Figure 3b gives the magnitude of the electric field along that axis in terms of the maximum magnitude E_m at the disk surface. It will be a slightly messy piecewise affair, but each component is simple. An annular disc has inner and outer radius R 1 and R 2 respectively. ), Electric field due to a charged infinite conducting plate, Simple Electric Field due to a Charged Disk, Electric field strength at a point due to 3 charges, Finding Area of Ring Segment to Find Electric Field of Disk, Electric field due to three point charges, 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. A couple of reminders: 1. It depends on the surface charge density of the disc. 2003-2022 Chegg Inc. All rights reserved. Electric Field Intensity due to continuous charge distribution | 12th physics |unit 01 Electrostatics |chapter 01Here in this video we are going to discuss a. Note that dA = 2rdr d A = 2 r d r. We will then collect terms that are proportional to 1/ r 3 and ignore terms thatare proportional to 1/ r 5 , where r = +( x 2 + y 2 )1 2 .We begin with33, Copyright 2020 123Doc. The deflection y2 isy2 = L2 tan 1 =eE y L1 L2mv0 2and the total deflection becomes21 eE y L1 eE y L1 L2 eE y L1 1y = y1 + y2 =+=L1 + L2 222 mv0mv0 22 mv02.13.4 Electric Field of a DipoleConsider the electric dipole moment shown in Figure 2.7.1. Yeah, but that's the problem. My attempt at a solution is shown in attached file "work for #10.png". Figure 25.15 shows one such ring. (b) What is the charge on the oil drop in units of electronic charge e = 1.6 1019 C ?Solutions:(a) The mass density oil times the volume of the oil drop will yield the total mass M ofthe oil drop,4M = oilV = oil r 3 3where the oil drop is assumed to be a sphere of radius r with volume V = 4 r 3 / 3 .Now we can substitute our numerical values into our symbolic expression for the mass,294 4 6314M = oil r 3 = (8.51 102 kg m 3 ) (1.6410 m) = 1.5710 kg33(b) The oil drop will be in static equilibrium when the gravitational force exactly balancesthe electrical force: Fg + Fe = 0 . @EmilioPisanty My best guess: Taking the origin to be at $O$, we have a system that can be thought of as a superposition of i) an infinite slab (infinite sheet with a finite thickness) of positive charge and ii) a sphere of radius $R$ centered at $O$ that is negatively charged. 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. The x component of the electric field strength at the point P with Cartesian coordinates ( x, y, 0)is given byEx =q cos + cos q xx=r 2 4 0 x 2 + ( y a) 2 3/ 2 x 2 + ( y + a ) 2 3/ 2 4 0 r+ 2 wherer 2 = r 2 + a 2 2ra cos = x 2 + ( y a ) 2Similarly, the y -component is given byEy =q sin + sin q yay+a=r 2 4 0 x 2 + ( y a) 2 3/ 2 x 2 + ( y + a) 2 3/ 2 4 0 r+ 2 We shall make a polynomial expansion for the electric field using the Taylor-seriesexpansion. For the former, we apply the superpositionprinciple:E=14 0qiri2riiFor the latter, we must evaluate the vector integralE=14 0dqrr2where r is the distance from dq to the field point P and r is the corresponding unitvector. What is. Why do we use perturbative series if they don't converge? Physics Galaxy. Okay, So for this particular problem we're talking about a charge electric charge distributed over X squared plus y squared, um, being less than or equal toe one. Complete step by step answer: Sinceobjects like planets have about the same number of protons as electrons, they areessentially electrically neutral. (e) What is the vertical displacement of the electron after time t1 when it leaves theplates? (d) The electron has an initial horizontal velocity, v 0 = v0 i . JavaScript is disabled. Relevant Equations:: Electric field due to disk. (3) Substitute dq into the expression for dE . Use MathJax to format equations. Where does the idea of selling dragon parts come from? 121 06 : 07. What is the velocity of theelectron at time t1 when it leaves the plates? But isn't having to calculate the electric field at any point in space, which in this case would be a suitable superposition of the previous two cases, a bit too much. The z axis scale is set by z_x= 8.0 cm. I cant see how you can go beyond setting up a good coordinate system and writing out a genralised integeral in terms of the parameters known. The slab carries a uniform charge density $\rho$ with the exception of a circular cavity that is carved out from the slab. of opposite charge. Select all correct statement(s) on the electric field, E when the charged disk is enormous (R -> ) or the point of interest is very close to the disk (z -> 0)? Thus,from the given data we can assert that there are five electrons on the oil drop!2.13.3 Charge Moving Perpendicularly to an Electric FieldAn electron is injected horizontally into a uniform field produced by two oppositelycharged plates, as shown in Figure 2.13.1. Asking for help, clarification, or responding to other answers. (b) Show that the above expression for the electric field can also be written in terms ofthe polar coordinates asE(r , ) = Er r + E whereEr =2 p cos p sin , E =34 0 r4 0 r 3Solutions:(a) Lets compute the electric field strength at a distance r a due to the dipole. (Notice that the term x / | x | only gives you the direction of the field, but doesn't change its magnitude.) Assertion :A uniformly charged disc has a pin hole at its centre. The oil drop has an unknown electric charge q (due toirradiation by bursts of X-rays). This problem has been solved! The z axis scale is set by z_x= 8.0 cm. also electric field at the centre . Xem v ti ngay bn y ca ti liu ti y (29.25 MB, 2,361 trang ), The above equation may be rewritten as z,1 2z + R2 2 0 Ez = z 1, 2z 2 + R2 0z>0(2.10.17)z<0The electric field Ez / E0 ( E0 = / 2 0 ) as a function of z / R is shown in Figure 2.10.9.Figure 2.10.9 Electric field of a non-conducting plane of uniform charge density.To show that the point-charge limit is recovered for zTaylor-series expansion: R2 1= 1 1 + 2 z z 2 + R2z1/ 2 1 R2= 1 1 +2 2 zR , we make use of the 1 R22 2 z(2.10.18)This gives R21 R 21 Q==Ez =222 0 2 z4 0 z4 0 z 2(2.10.19)which is indeed the expected point-charge result. Q. P.S: Setting O as the origin (arbitrary point on the cylinder's central axis) the answer is rho*R/2epsilon knot . The magnitude of the charge of the electron and proton is. E=k2[1 z 2+R 2z] where k= 4 01 and is the surface charge density. For a point charge, the potential V is related to the distance r from the charge q, V = 1 4 0 q r. a. Homework Statement. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. What properties should my fictional HEAT rounds have to punch through heavy armor and ERA? Notice that we have chosen the unit vector j to point upward. Step 2 - Permittivity of Free Space (Eo) Step 3 - Enter the Radius. Then why are the large scale motions of planetsdetermined by the gravitational force and not the electrical force. For a better experience, please enable JavaScript in your browser before proceeding. Japanese girlfriend visiting me in Canada - questions at border control? In finding the electric field due to a thin disk of charge, we use the known result of the field due to a ring of charge and then . You are using an out of date browser. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. (b) The acceleration of the electron isa=qEeEqE= y j = y jmmmand its direction is upward.31(c) The time of passage for the electron is given by t1 = L1 / v0 . (6) Complete the integration to obtain E .In the Table below we illustrate how the above methodologies can be utilized to computethe electric field for an infinite line charge, a ring of charge and a uniformly charged disk.Line chargeRing of chargeUniformly charged diskdq = dxdq = ddq = dAFigure(2) Express dq interms of chargedensity(3) Write down dE(4) Rewrite r and thedifferential elementin terms of theappropriatecoordinates(5) Apply symmetryargument to identifynon-vanishingcomponent(s) of dEdE = ker2dE = kedxcos =yr ydx( x + y )2+ /2 /2=rdE = ke22 3/ 2r = R2 + z2dx( x2 + y2 )3/ 2/22key ( / 2)2 + y2(R + z )22 3/ 2R z( R + z 2 )3/ 2(2 R ) z= ke 2 2 3/ 2(R + z )Qz= ke 2 2 3/ 2(R + z )E z = ke2r2dEz = dE cos Rz d = ke dAdA = 2 r ' dr 'zcos =r2r = r + z 2dEz = dE cos dE y = dE cos = ke dld = R d zcos =rr = x2 + y 2Ey = ke y(6) Integrate to get E dx d = ke2 zr dr (r 2 + z 2 )3/ 2Ez = 2ke zR0r dr(r + z2 )3/22 zz = 2ke 22| z | z + R 272.13Solved Problems2.13.1 Hydrogen AtomIn the classical model of the hydrogen atom, the electron revolves around the proton witha radius of r = 0.53 10 10 m . By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. you can sum many dr to make a disk de-ce (1-3 Z ZARZ 2 edrz JE = LITE (2 +19) /4 E= SR cezx 25 E = 2E0 Point charge in electric field 7rq Dipole in an electric field. Yes, I know how to compute the $E$ field due to an infinite slab -- infinite with a finite thickness. Note that the motion of the electron is analogous to the motion of a mass that isthrown horizontally in a constant gravitational field. Would it be possible, given current technology, ten years, and an infinite amount of money, to construct a 7,000 foot (2200 meter) aircraft carrier? (ii) In constant electric field along z-direction, the perpendicular distance between equipotential surfaces remains same. What is the total vertical displacement of the electron from time t = 0 until it hits thescreen at t2 ?Solutions:(a) Since the electron has a negative charge, q = e , the force on the electron isF e = qE = eE = (e)( E y )j = eE y jwhere the electric field is written as E = E y j , with E y > 0 . The Ionospheric Photometer (IPM) instrument onboard the FY-3(D) meteorological satellite was employed to . I used Desmos Scientific online calculator to obtain my final answer. Since the gravitational force points downward, theelectric force on the oil must be upward. Okay, then we're saying that the charge is has a density such that we have a function equaling the square root of X squared plus y squared. (c) The plates have length L1 in the x -direction. $$S_2=\{(x,y,z)\in\mathbb R^3: -dR^2\}.$$. How does legislative oversight work in Switzerland when there is technically no "opposition" in parliament? Does the result depend on the distance between the proton and theelectron? which is the expression for a field due to a point charge. I get the same answer as you, using the formula provided. How could my characters be tricked into thinking they are on Mars? The sphere will have its $E$ field in the radial direction and the slab will have its $E$ field in the $z$ direction. Magnetic field of a rotating disk with a non-uniform volume charge. Next week (February 1-5), there will be a Lab quiz on the concepts covered in labs one and two Lecture notes: The Electric Field Due to a Charged Disk Charge per unit area = , therefore the total amount go charge in a ring of radius r and width dr is dq = dA = (2 r dr) The contribution to the electric field due to this ring . A uniformly charged disk of radius 35.0 cm carries charge with a density of 7.90 x 10^-3 C/m^2. (4) Specify an appropriate coordinate system (Cartesian, cylindrical or spherical) andexpress the differential element ( d , dA or dV ) and r in terms of the coordinates (seeTable 2.1 below for summary. 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. helps visualize this configuration: Find the electric field everywhere in space. The Electric Field Due to a Charged Disk Figure 3a shows a circular disk that is uniformly charged. Therefore the force between planets is entirely determinedby gravity.2.13.2 Millikan Oil-Drop ExperimentAn oil drop of radius r = 1.64 106 m and mass density oil = 8.51 102 kg m3 isallowed to fall from rest and then enters into a region of constant external field E appliedin the downward direction. If you are just looking for a list of demos, the navigator on the left side of the screen includes a categorized listing of all of the demos currently owned by the Department of Physics at Indiana University. How to use Electric Field of Disk Calculator? (d) The electric force is 39 orders of magnitude stronger than the gravitational forcebetween the electron and the proton. To do this, simply superpose the field from a thick slab (easy) with the field from an oppositely charged sphere (easy). Electric field generated by disk and conductor, Electric field on rim of uniformly charged disk, Gauss Law to Find Electrical Field on central axis from Disk of Uniform Charge Density, Magnetic field induced by a rotating charged disk, Application of Gauss Law to Find the Electric Field for an Arbitrary Point in a Ring of Charge in 2D (and Similar Problems). The electric field at the centre of the disc is zero Reason: Disc can be supported to be made up of many rings. Denote the distance along the z axis from the center of the disk (O) to the point P (on the z axis) by z. a radius of r = 0.53 10 10 m . What have you tried so far? identify non-vanishing component (s) of the electric field. You'll get a detailed solution from a subject matter expert that helps you learn core concepts. The OI135.6 nm radiation intensity and the associated change with solar activity are very complex, and this is particularly the case during November 2020. Find the electric field due to a Charged Disk at a distance of "d" which is in the disk's axis direction. Isn't this kind of a hopeless task? Thank you. which can be solved exactly (as long as $R0(2.10.20)z<023The plot of the electric field in this limit is shown in Figure 2.10.10.Figure 2.10.10 Electric field of an infinitely large non-conducting plane.Notice the discontinuity in electric field as we cross the plane. Why? Does illicit payments qualify as transaction costs? (d) In light of your calculation in (b), explain why electrical forces do not influence themotion of planets.Solutions:(a) The magnitude of the force is given by1 e2Fe =4 0 r 2Now we can substitute our numerical values and find that the magnitude of the forcebetween the proton and the electron in the hydrogen atom isFe =(9.0 109 N m 2 / C2 )(1.6 1019 C)2= 8.2 108 N112(5.3 10 m)(b) The magnitude of the electric field due to the proton is given byE=q (9.0 109 N m 2 / C2 )(1.6 1019 C)== 5.76 1011 N / C4 0 r 2(0.5 1010 m) 21(c) The mass of the electron is me = 9.1 10 31 kg and the mass of the proton ism p = 1.7 1027 kg . For lesser than 2R and further lesser than R, you follow the same method. Gauss' law comes in. The Electric Field Due to a Charged Disk Question 10: The electric field due to a thin spherical shell having a charge 'q', is given as _____, where 'r' is the distance of the point from the center of the shell, (outside the shell). (b) What is the magnitude of the electric field due to the proton at r? What is this fallacy: Perfection is impossible, therefore imperfection should be overlooked, Finding the original ODE using a solution. Suggested for: Electric field due to a charged disk. P.SL We haven't done special functions in the course uptil now so I guess no one really expects us to use those in this problem. How can a region of uniform charge density have an an axial (parallel to only one axis) electrostatic field? Physically this means that the plane is very large, or thefield point P is extremely close to the surface of the plane. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Equipotential surface is a surface which has equal potential at every Point on it. 6 10 19 C , the charge of the oil drop in units of e isN=q 8.02 1019 C==5e 1.6 1019 CYou may at first be surprised that this number is an integer, but the Millikan oil dropexperiment was the first direct experimental evidence that charge is quantized. I have the solution for uniformly charged disk but I can't figured it out for the situation above. On the other hand, we may alsoconsider the limit where R z . Consider the electric field due to a point charge Q Q size 12{Q} {}. No one would help you if you don't show any attempt. Figure 3b gives the magnitude of the electric field along that axis in terms of the maximum magnitude E_m at the disk surface. For example, related to the problem of a a unifiromly charged disk, Purcell and Morin's textbool Electricity and Magnetism reads: I have been given the following question: Consider a slab of thickness $2R$ that extends to infinity along the other two dimensions. Surface charge density is .Find the electric field at any point distant y along the axis of the disc. which is easy enough, may be instructive. (' o ' is the permittivity of free space) CGAC2022 Day 10: Help Santa sort presents! The charge has it's maximum value at disk's center and decreases towards the edges. Step 4 - Enter the Axis. The discontinuity is givenbyEz = Ez + Ez = =2 0 2 0 0(2.10.21)As we shall see in Chapter 4, if a given surface has a charge density , then the normalcomponent of the electric field across that surface always exhibits a discontinuity withEn = / 0 .2.11 SummaryThe electric force exerted by a charge q1 on a second charge q2 is given byCoulombs law:F12 = keq1q21 q1q2r =r2r4 0 r 2whereke =14 0= 8.99 109 N m 2 / C2is the Coulomb constant.The electric field at a point in space is defined as the electric force acting on a testcharge q0 divided by q0 :Feq0 0 q0E = lim24The electric field at a distance r from a charge q isE=qr4 0 r 2Using the superposition principle, the electric field due to a collection of pointcharges, each having charge qi and located at a distance ri away isE=1r4 0i2riiA particle of mass m and charge q moving in an electric field E has an accelerationa=qi1qEmAn electric dipole consists of two equal but opposite charges. A charge distributed uniformly over a disc will produce an electric field. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. If we bring a charged particle from infinity to a point in this field, we need to do some work. To learn more, see our tips on writing great answers. (d) Suppose the electron enters the electric field at time t = 0 . Convert $\hat{r}$ to Cartesian components and add. I also know how to compute the potential due to a uniformly charged disk on the symmetry axis. the electric field for an infinite line charge, a ring of charge and a uniformly charged disk. We will use a ring with a radius R' and a width dR' as charge element to calculate the electric field due to the disk at a point P . These functions are Is the electric field at the edge of a uniformly charged disk infinite? We review their content and use your feedback to keep the quality high. (g) The electron hits the screen located a distance L2 from the end of the plates at a timet2 . Spherical cavity is easy and can be related to the gravitational field calculations (after adjusting constants and stuff) and even for cylinders it will be easy by considering a cylinder to be a wire with lambda as the charge density. This is at odds with the question statement but it usefully narrows down the set in question to The mass follows a parabolictrajectory downward. Problem: Consider a disk of radius R with a uniform charge density . The. Using the law derive an expression for electric field due to a uniformly charged thin spherical shell at a point outside the shell. (b) What is the acceleration of the electron when it is between the plates? It is a hopeless task to calculate the field of this thickened disk anywhere but on its axis of symmetry - and certainly not without some very significant involvement of special functions. The force on the electron isupward. The central z axis is perpendicular to the disk face, with the origin at the disk. How should I go about the problem? @junaid If the cavity is spherical then the calculation is trivial. Making statements based on opinion; back them up with references or personal experience. Hint: Suppose if a circular disc has a surface charge density, it will produce an electric field along the axis.The field strength varies as we go from the surface to a point in the axis. 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