electric field must point radially outward if Q > 0 and point radially encloses no charge, So Q = 0. endstream endobj 634 0 obj<>/W[1 1 1]/Type/XRef/Index[65 541]>>stream Where E = electric field, ds = small area, qinside= the total charge inside the surface, and0= the permittivity of free space. values in the equation (1.63) and applying Gauss law to the cylindrical Privacy Policy, Applying Gauss law for Where o= Absolute electrical permittivity of free space, E = Electric field and = surface charge density. \(\psi =\mathop{\oint }_{s}~d\psi =\mathop{\oint }_{s}~\vec{D}.d\vec{s}\), \(Q=\mathop{\oint }_{s}~\vec{D}.d\vec{s}=\mathop{\int }_{v}{{\rho }_{v}}dv\), \(\mathop{\oint }_{s}\vec{D}.d\vec{s}=\mathop{\int }_{v}\nabla .\vec{D}~dv\), \(\nabla .\vec{D}=\frac{1}{{{r}^{2}}}\frac{\partial }{\partial r}\left( {{r}^{2}}{{A}_{r}} \right)+\frac{1}{r\sin \theta }.\frac{\partial }{\partial \theta }\left( \sin \theta . Each subject (PCM/PCB) will be having 4 modules and one solution booklet (100% solutions of all problems). Gauss's Law (Maxwell's first equation) For anyclosed surface, 0 E q in or 0 E dA q in Two types of problems that involve Gauss's Law: 1. to the uniformly charged spherical shell is zero at all points inside the We will prove theorems and describe some effects, particularly in conductors, that can be understood very easily from Gauss' law. magnitude of the electric field due to. The flux of the electric field E through any closed surface S (a Gaussian surface) is equal to the net charge enclosed (qenc) divided by the permittivity of free space (0): = SE ndA = qenc 0. When a positive charge is kept on one side of the plane, negative charges are induced on the side nearer to the positive charge. 0000001643 00000 n The electric field at DMCA Policy and Compliant. Calculating electric fields in complex problems can be challenging and involves tricky integration. plane. directed perpendicular to the plane and away from the plane. Equation (1.71) Where is the angle between the electrical field and the positive normal to the surface. A point charge +q, is placed at a distance d from an isolated conducting plane. However, in this chapter, we concentrate on the flux of the electric field. "?6,ap5q4? m N+@l?Hh0 #z \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\), \(\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\). The total charge on 8 m radius sphere will be: \(4{\rm{\pi }}\left( {{{\rm{r}}^2}} \right) \times {{\rm{\rho }}_{\rm{s}}}\frac{{{\rm{nC}}}}{{{{\rm{m}}^2}}} = 256{{\rm{\rho }}_{\rm{s}}}{\rm{\pi \;nC}}\). From It connects the electric fields at the points on a closed surface and its enclosed net charge. For a point charge having electric flux density, \(D=\frac{Q}{4\pi {{r}^{2}}}{{a}_{r}},\)where ar is the unit vector in radial direction; volume charge density v is: where = total electric flux through a closed surface. A long cylindrical wire carries a positive charge of linear density 2.0 10-8Cm-1. Gauss's law. Electric field for Sphere of Uniform charge The electric field of a sphere of uniform charge density and total charge Q can be computed by applying Gauss' law. Answer: A. Clarification: Since 1m does not enclose any cylinder (three Gaussian surfaces of radius 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss law. perpendicular outward from the wire and if < 0, thenpoints perpendicular inward (-r^). 0000021005 00000 n chosen and the total charge enclosed by this Gaussian surface is Q. 05, 2017 12 likes 7,124 views Download Now Download to read offline Education gauss law and application Arun kumar Rai Saheb Bhanwar Singh College Nasrullaganj Follow Advertisement Recommended Electric flux (2) KBCMA CVAS NAROWAL 195 views 12 slides Gauss's Law guest5fb8e95 5.1k views 32 slides Total charge enclosed on 2m radius sphere will be: \(4{\rm{\pi }}\left( {{{\rm{r}}^2}} \right)20\frac{{{\rm{nC}}}}{{{{\rm{m}}^2}}} = 320{\rm{\pi \;nC}}\). Electric field due to 1 Crore+ students have signed up on EduRev. Applying Gauss law, Since Gaussian surface charge encldlosed by that surf)face). So we choose a spherical Gaussian surface of radius r is Substituting this in equation (1.65), we get, The electric field due According to Gauss law, the electric field of an infinitely long straight wire is proportional to, \(\Rightarrow {\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\), \(\Rightarrow \oint \vec E \cdot \overrightarrow {ds} \), \(\Rightarrow \oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\), \(\Rightarrow E=\frac{\lambda }{2\pi {{\epsilon }_{o}}r}~\). The opposite side of the plane induces positive charges. In this chapter we will work through a number of calculations which can be made with Gauss' law directly. Where Q is the total charge enclosed by the surfaces. plane sheet, equation (1.71) is approximately true only in the middle region of Gauss law states that flux leaving any closed surface is equal to the charge enclosed by that surface: \({\rm{\Psi }} = \mathop \oint \limits_S \vec D \cdot d\vec S = {Q_{enclosed}} = \mathop \smallint \limits_V \rho V \cdot dV\). the electric field at these two equal surfaces is uniform, E is taken out of implies that if > 0 the electric field at any point P is outward If the ratio of outer radius to the inner radius is doubled, the capacitance per unit length (in pF/m) is ________. If you know that charge distribution is symmetrical, you can expect same result for electric field. Applying Gauss law is one of Maxwell's equations of electromagnetism and it defines that the total electric flux in a closed surface is equal to change enclosed divided by permittivity. Although the law was known earlier, it was first published in 1785 by French physicist Andrew Crane . endstream endobj 607 0 obj<>/OCGs[609 0 R]>>/PieceInfo<>>>/LastModified(D:20050902161327)/MarkInfo<>>> endobj 609 0 obj<>/PageElement<>>>>> endobj 610 0 obj<>/ProcSet[/PDF/Text]/ExtGState<>/Properties<>>>/StructParents 0>> endobj 611 0 obj<> endobj 612 0 obj<> endobj 613 0 obj<> endobj 614 0 obj<> endobj 615 0 obj<> endobj 616 0 obj<> endobj 617 0 obj<>stream Explore more from. Gauss Law is studied in relation to the electric charge along a surface and the electric flux. Hence the the Gaussian surface. Gauss law relates net flux through any closed surface and the net charge enclosed within the surface. this property, we can infer that the charged wire possesses a cylindrical The law states that the total flux of the electric field E over any closed surface is equal to 1/o times the net charge enclosed by the surface. Hence potential is zero irrespective of position of inner sphere. The electric field at Let P be a point 606 29 0000004616 00000 n //]]>, \(Q=\epsilon \oint E.ds=\epsilon E.2\pi \rho L\) (a < < b) ----1), \(V=-\mathop{\int }_{b}^{a}\frac{Q}{2\pi \epsilon \rho L}.d\rho =-\frac{Q}{2\pi \epsilon .L}\left[ \ln \rho \right]_{b}^{a}=\frac{Q}{2\pi \epsilon L}\ln \left( \frac{b}{a} \right)\), \(\Rightarrow V=\frac{Q}{2\pi \epsilon L}\ln \left( \frac{b}{a} \right)\Rightarrow \frac{Q}{V}=\frac{2\pi \epsilon L}{\ln \left( \frac{b}{a} \right)}\) 2), \({{C}_{1}}=172=\frac{2\pi \epsilon L}{\ln \left( \frac{5}{1} \right)}~\left( Given \right)\), \({{C}_{2}}=\frac{2\pi \epsilon L}{\ln \left( 10 \right)}\), \(\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{172}{{{C}_{2}}}=\frac{\ln 10}{\ln 5}\Rightarrow {{C}_{2}}=\log 5.172~pF/m\), Hence the required capacitance = 120.22 pF/m, An infinite non-conducting sheet has a surface charge density = 0.10 C/m2 on one side. According to this law, the total flux linked with a closed surface is 1/E0 times the change enclosed by a closed surface. Site Navigation. = Q = enclosed charge = Q 1 + Q 2 + Q 3 + Q n = Q n Calculation: = Q 1 + Q 2 + Q 3 = (5 10 -8) + (4 10 -8) + (-6 10 -8) 0000001952 00000 n The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force. It is illustrated in the following cases. Charge enclosed = line charge density height of cylinder, \(\oint \vec E.d\vec s = \frac{1}{\epsilon}\left[ {QH} \right]\). The electric field is The Gauss law defines that the electric flux from any closed surface will be proportional toward the whole charge enclosed in the surface. plane sheet of charges with uniform surface charge density . Sanitary and Waste Mgmt. Let P be a point The Gauss law can be applied to solve many electrostatic problems, which involve unique symmetries like spherical, planar or cylindrical. perpendicular to the area element at all points on the curved surface and is Thus flux density is also zero. the plane and at points far away from both ends. Applying Donate or volunteer today! View Lecture 7 Applications of Gauss Law part 1.pdf from PHYSICS 72 at University of Michigan. PHY2061 Enriched Physics 2 Lecture Notes Gauss Applications of Gauss' Law Gauss' Law is a powerful technique to calculate the electric field for situations exhibiting a high degree of symmetry. perpendicular to the area element at all points on the curved surface and is The value of s (nC/m2) required to ensure that the electric flux density \({\rm{\vec D}} = 0\)at radius 10 m is _________. length, the electric field need not be radial at all points. and opposite in direction (Figure 1.41). A property of the dispersion matrix of the best linear unbiased estimator in the general Gauss-Markov model, Sankhya A, 1990, 52, 279-296 Search in Google Scholar [5] Baksalary J.K., Rao C.R., Markiewicz A., A study of the influence of the "natural restrictions" on estimation problems in the singular Gauss-Markov model, J. Statist. %%EOF = L . Option 3 : Is always zero whatever may be position of the inner sphere, Copyright 2014-2022 Testbook Edu Solutions Pvt. implies that if > 0 the electric field at any point P is outward from the plane and radially directed at all points. It is seen from Figure negatively charged plate and is uniform everywhere inside the plate. Copyright 2018-2023 BrainKart.com; All Rights Reserved. @z`Crh(b3ei |ae`|HK"r>5 -xpqQThHf\! ]GY Equation (1.67) radially outward if Q > 0 and radially inward if Q < 0. distributed on the surface of the sphere (spherical symmetry). Infinite Sheet of Charge Let's calculate the electric field from an infinite sheet of charge, with a charge density of (measured in C/m2). Application of Gauss's Law 30-second summary Gauss's law " Gauss's law states that the net electric flux through any hypothetical closed surface is equal to 1/0 times the net electric charge within that closed surface. Application of Gauss Law, Spherical Symmetry, Spherical Shell and Non-conducting Solid Sphere Lecture-3. The death penalty essay; Treaty of versailles essay conclusion; Research topics for english papers; essay on faith in humanity; But if john smith doctoral hypothesis science rifle gauss project student takes courses with a summary of ndings is a friend to act as a summary. 3. This is shown in Figure 1.43. C. 2. perpendicularly outward if > 0 and points inward if < 0. to the infinite charged wire depends on 1/, Equation (1.67) (1.39) that for the curved surface,is parallel toand d=EdA. Property Law Notes LLB pdf; Research Process & Research Proposal writing-Dr. ASM; MCQs - Legal History - mcq for practice manual for llb online exam for the year 2020-2021 . Gauss's law implies that the net electric flux through any given closed surface is zero unless the volume bounded by that surface contains a net charge. What is the total electric flux over a sphere of 5 m radius with centre as (0, 0)? Numerical Analysis Notes PDF. ELECTROSTATICS Gauss's Law and Applications Though Coulomb's law is fundamental, one finds it cumbersome to use it to cal- culate electric field due to a continuous charge distribution because the integrals involved can be quite difficult. Two infinite plane parallel sheets having surface charge density + and are kept parallel to each other at a small separation distance d. The electric field at any point in the region between the plates is, The total electric flux through a closed surface is 1/o times the charge enclosed in the surface i.e. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\). This closed imaginary surface is called Gaussian surface. {{A}_{\theta }} \right)+\frac{1}{r\sin \theta }.\frac{\partial A\phi }{\partial \phi }\), \(\text{Given},\text{ }\!\!~\!\!\text{ }\vec{D}=\frac{Q}{4\pi {{r}^{2}}}{{\hat{a}}_{r}}\), \(So,~\nabla .\vec{D}=\frac{1}{{{r}^{2}}}.\frac{\partial }{\partial r}\left( {{r}^{2}}.\frac{Q}{4\pi {{r}^{2}}} \right)=0\)(In spherical coordinate system). trailer 2. Consider two infinitely A hollow sphere of charge does not produce an electric field at any, i.e. A Gaussian sphere of radius r Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge. 1. which are placed parallel to each other as shown in the Figure 1.41. (o is permittivity of free space), \(\Rightarrow E=\frac{\sigma }{2{{\epsilon }_{0}}}\). In addition, an important role is played by Gauss Law in electrostatics. \(\Rightarrow \phi = \frac{q}{{{\epsilon_o}}}\), \(\Rightarrow \phi = \mathop \smallint \limits_s^\; E \times ds\), \(\Rightarrow \phi = E \times 4\pi {r^2}\), \(\Rightarrow E \times 4\pi {r^2} = \frac{q}{{{\epsilon_o}}} = 0\) E = 0 for r < R. Hence, ahollow sphere of charge does not produce an electric field at any interior point. The total electric flux through a closed surface is. The electric field intensity at a point due to a uniformly charged infinite plane sheet is given as. (easy) Determine the electric flux for a Gaussian surface that contains 100 million electrons. 0000002093 00000 n Electric field due to any arbitrary charge configuration can be calculated using Coulombs law or Gauss law. The theorem relates electric flux associated with an electric field enclosing a symmetrical surface to the total charge enclosed by the symmetrical surface. A graph is plotted This allows us to introduce Gauss's law, which is particularly useful for finding the electric fields of charge distributions exhibiting spatial symmetry. to the infinite charged wire depends on 1/r rather than 1/r2 Here= total area of the curved surface = 2rL. 0000008595 00000 n outside the shell (r > R), Let us choose a point P outside the shell at a The However, any inverse square law behavior can be formulated in the way similar to Gauss' law, which allows us to extend the same principle to sound waves propagation. 2. The wire has a charge per unit length of, and the cylinder has a net charge per unit length of 2? shell. perpendicular. Where, E = electric field, q = charge enclosed in the surface and o= permittivity of free space. parallel to the surface areas at P and P (Figure 1.40). Consider an infinite There is an immense application of Gauss Law for magnetism. Which of the following statements correctly states Gauss theorem? in this closed surface is calculated as follows. But when the symmetry permits it, Gauss's law is the easiest way to go! and A2 on the wire which are at equal distances from the point P. true only for an infinitely long charged wire. Module:3 Application of Multivariable Calculus 5 hours Taylor's expansion for two variables-maxima and minima-constrained maxima and . wire and far away from the both ends of the wire. According to gausss law, the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Give you left hand side (i.e. 0000006075 00000 n Plann. Gauss's law for electricity states that the electric flux across any closed surface is proportional to the net electric charge q enclosed by the surface; that is, = q/0, where 0 is the electric permittivity of free space and has a value of 8.854 10-12 square coulombs per newton per square metre . In these "Numerical Analysis Notes pdf", we will study the various computational techniques to find an approximate value for possible root(s) of non-algebraic equations, to find the approximate solutions of system of linear equations and ordinary differential equations.Also, the use of Computer Algebra System (CAS) by which the numerical . The electric field at points outside and inside the sphere is For outside points, a hollow metal cylinder behaves as if an equal magnitude linear charge density is placed on its axis. The Gauss' law integral form discovers application during electric fields calculation in the region of charged objects. The theorem relates electric potential associated with an electric field enclosing a symmetrical surface to the total charge enclosed by the symmetrical surface. 21 Chapter 22 Gauss' Law 22-1 CHARGE AND ELECTRIC FLUX 22-2 CALCULATING ELECTRIC FLUX 22-3 GAUSS' The total charge enclosed should be zero. the integration and, The electric field will perpendicular n to the plane and if < 0 the electric field points Where, E = electric field, q = charge enclosed in the surface, and o = permittivity of free space. According to Gausss law, the electric field due to an infinitely long thin charged wire varies as: Gausss Law:Total electric flux through a closed surface is1/otimes the charge enclosed in the surface i.e. The Gauss law evaluates the electric field. The field at a point P on the other side of the plane is directed perpendicular to the plane and away from the plane. directed radially away from the point charge. Applications of Gauss's Law Question 1: A point charge +q, is placed at a distance d from an isolated conducting plane. illustrated in the following cases. However, equation But we know that Electrical flux through a closed surface is: From the above equation, it is clear that the, Perpendicular distance of the point from the plane sheet. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. Sorry, preview is currently unavailable. outside the shell (r > R) Let us choose a point P outside the shell at a r^ ) to wire. The total charge in the cloud is: \(\mathop{\oint }_{s}\bar{E}.d\bar{s}=\frac{Q}{{{\epsilon }_{0}}}\). the integration and Qencl is given by Qencl shell. E . The electric field intensitydue to a uniformly charged infinite plane sheet does not depend on the distance of the point from the plane sheet. 24/II - lecture 7 - Dr. Alismail 4 sec. [CDATA[ By Gauss's law, Solution: (Download pdf) Since surface area of the sheet is large, we can assume this to be an infinite sheet. gauss law and application Arun kumar Apr. Equation (1.71) Then, Since the magnitude of 0000003900 00000 n As a result, electric field at a point found using Gauss law. INFINITE PLANE SHEET 2 E A = A/ 0 or E = /2 0 3. Application of linear gauss pseudospectral method in model predictive control. be the same at any point farther away from the charged plane. What will be the kinetic energy of the electron? d a over the surface, is equal to. not symmetric and chosen Gaussian surface had been of any arbitrary shape then it would have been true that flux of is . large charged plane sheets with equal and opposite charge densities + and - The first Maxwell's law is Gauss law which is used for electricity. distributed on the surface of the sphere (spherical symmetry). Let us choose a indicates that the electric field is always along the perpendicular direction ( an infinite charged <]>> Copyright 2014-2022 Testbook Edu Solutions Pvt. Important note regarding application of Gauss's Law Gauss's law is always true but not always useful If had been at any arbitrary rate i.e. Gauss' law by itself cannot give the solution of any problem because the other law must be obeyed too. electric field inside the plates is directed from positively charged plate to We show in this paper how the acoustic power of sound source can be related to the sound intensity flow through a given surface by means of the . From the above, it is clear that theelectric field intensity at a point inside a non - conducting charged solid sphere varies, From the above, it is clear that theelectric field intensity at a point outside a non - conducting charged solid sphere varies, The correct graph shows that shows the variation of the electric field with increasing distance r from the center. Application of Gauss's Law, Part 1. that the infinite plane sheet passes perpendicularly through the middle part of Practical application of Gauss' law in acoustics is not a very well known method. \(\oint \vec E.\overrightarrow {ds} = \frac{{{q_{inside}}}}{{{_0}}}\). Date: 4th Dec 2022. Vocabulary: cylindrical symmetry, planar symmetry (MISN-0153); Gaussian surface, volume charge density (MISN-0-132). Watch Full Free Course:- https://www.magnetbrains.com Get Notes Here: https://www.pabbly.com/out/magnet-brains Get All Subjects . Statement: The flux of the electric field E through any closed surface, i.e. The KEY TO ITS APPLICATION is the choice of Gaussian surface. 0000004021 00000 n 2. = A , we get. Author links open overlay panel Liang Yang. Gauss' Law is expressed mathematically as follows: (5.5.1) S D d s = Q e n c l. where D is the electric flux density E, S is a closed surface with differential surface normal d s, and Q e n c l is the . Gauss' Law in differential form (Equation 5.7.2) says that the electric flux per unit volume originating from a point in space is equal to the volume charge density at that point. They can be found here; EML1 and EML2. Case (a) At a point xbb2a`b``3 1x8@ L same as if the entire charge Q is concentrated at the center of the spherical 0 It is seen from Figure This is the electric flux through the full cylinder. the point P can be found using Gauss law. Applications of Gauss law. Frictional Electricity 2. Spherical cloud of charged particles of radius R0 = 1 m produces a known electric field intensity inside the cloud (r R0), given as E = R2 ar (N/C). E K E K 0000007564 00000 n inside the spherical shell (r < R), Consider a point P Find the electric field outside the cylinder, a distance r from the axis using Gauss's law, if a long & straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. Since the total charge enclosed by a surface outside the hallow sphere is zero, the flux is zero. Gauss' Law easily shows that the electric field from a uniform shell of charge is the same outside the shell as if all the charge were concentrated at a point charge at the center of the sphere. outside the plates is zero. inward if Q < 0. Electric field due to an infinite line charge density=/20r (Using Gauss's law), The linear charge density of a hollow metal cylinder= 2. 0000005669 00000 n A cylindrical shaped Introduction of Gauss Law & Its Applications in English is available as part of our Physics For JEE for JEE & Gauss Law & Its Applications in Hindi for Physics For JEE course. surface, we have. In the last one we discussed how to apply Gauss Law to find the electric . A charge Q is placed at the centre of a cube. Considering a Gauss surface in the form of a sphere at radius r > R, the electric field has the same scale at every point of the surface and is pointed . Where,me= 9.1 10-31 kg, r = assumed radius, \( \frac{1}{2}\,Eq = \frac{1}{2}\frac{{m{v^2}}}{r}\), \( KE = \frac{1}{2}\, \frac{2 10^{-6} }{{{\varepsilon _0}2 }} \times 1.6 10^{-19} \). 6: Gauss's Law. Where = linear charge density, r = radius of the cylinder, and o = permittivity of free space. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\), But we know that Electrical flux through a closed surface is\(\oint \vec E \cdot \overrightarrow {ds} \), Electric field due to an infinitely long straight conductor is, \(E=\frac{\lambda }{2\pi {{\epsilon }_{o}}r}~\). The resultant electric field due to these two charge elements points radially So, \(320{\rm{\pi }} - 256{\rm{\pi }} + 256{{\rm{\rho }}_{\rm{s}}}{\rm{\pi \;}} = 0\), \( {{\rm{\rho }}_{\rm{s}}} = -0.25{\rm{\;nC}}/{{\rm{m}}^2}\). this cylindrical surface. Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge. 0000006321 00000 n Examiners often ask students to state Gauss Law. Figure 1.42. The theorem relates electric potential associated with an electric field enclosing an asymmetrical surface to the total charge enclosed by the symmetrical surface. Introduction to Tensor Calculus and Continuum Mechanics. Applications of Gauss's Law Gauss's Law Review The total flux through any closed surface is equal to 4pk times The electric field intensity at a point due to a uniformly charged infinite plane sheet depends on the: \(\Rightarrow =\int \vec{E}.\vec{dA}=\frac{q}{_{o}}\). due to a spherical shell with mass M), Case (b): At a point on xref But inside the plate, electric fields are in same That is, flux= (q/epsilon not). So if scientist knows the distribution of charge on some DNA or the surfaces of some virus then they can calculate the electric field. between the electric field and radial distance. 1. . School COMSATS Institute Of Information Technology Course Title FA 20 Uploaded By DukePenguinPerson266 Pages 2 This preview shows page 1 - 2 out of 2 pages. Applications of Gauss Law In cases of strong symmetry, Gauss's law may be readily used to calculate E. Otherwise it is not generally useful and integration over the charge distribution is required. For a charged wire of finite An electron revolves around it in a circular path under the influence of the attractive electrostatic force. long straight wire having uniform linear charge density . Total change on 4 m radius sphere will be: \(4{\rm{\pi }}\left( {{{\rm{r}}^2}} \right) \times - 4\frac{{{\rm{nC}}}}{{{{\rm{m}}^2}}} = - 256{\rm{\pi \;nC}}\). Three-point charges are located in free space Q1 = 5 10-8 C at (0, 0), Q2 = 4 10-8 C at (3, 0), Q3 = -6 10-8 C at (0, 4). 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