why is symmetry useful when applying gauss's law?

It only takes a minute to sign up. Specifically, the integral for the flux is easiest to evaluate if: Ultimately, the points above should dictate the choice of gaussian surface so that the integral for the flux is easy to evaluate. Gauss Law is a relation between the net flux through a closed surface and the amount of charge, \(Q^{enc}\), in the volume enclosed by that surface: \[\oint\vec E\cdot d\vec A=\frac{Q^{enc}}{\epsilon_{0}}\]. Therefore, we set up the problem for charges in one spherical shell, say between and ,as shown inFigure 2.3.6. 0000003967 00000 n You might ask, why not the volume of the sphere? Choose a gaussian surface that goes through the point for which you want to know the electric field. Let be the area of the shaded surface on each side of the plane and be the magnitude of the electric field at point . S E da = Qenc 0. It is actually a triple integration that has to be integrated over a volume. In this example, we used Gauss Law to determine the electric field above an infinite plane. Let us choose a box (right panel of Figure \(\PageIndex{11}\)) of length, \(L\), with a square cross-section of side, \(a\). They are the only surfaces that give rise to nonzero flux because the electric field and the area vectors of the other faces are perpendicular to each other. 0000001536 00000 n Also, note that at the center of the sphere, the electric field has a magnitude of zero, as expected from symmetry. The charge enclosed by the Gaussian surface is given by. Next, we determine the magnitude of the electric field inside the charged sphere. 0000002652 00000 n Inside the sphere, we found that the electric field is zero at the center, and increases linearly with radius. Should convert 'k' and 't' sounds to 'g' and 'd' sounds when they follow 's' in a word for pronunciation? 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 how Pressbooks supports open publishing practices. Why is this so?? Although this is a situation where charge density in the full sphere is not uniform, the charge density function depends only on the distance from the centre and not on the direction. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Since for symmetrie reasons, the field should be the same for every point on the sphere and the direction of the field should be perpendicular on the surface, it becomes much easier: $$ d) It allows us to determine the strength of the electric field. Why, when and where is Gauss's law applicable? Similarly, if there are the same number of field lines entering a closed surface as there are lines exiting that surface (a flux of zero), then the surface encloses no charge. They are. To learn more, see our tips on writing great answers. Yes. 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. 0000004538 00000 n The determination of the electric field using Gauss's law requires determining the symmetry of a particular charge distribution and predicting the direction of the electric field everywhere if a high symmetry exists. CEO Update: Paving the road forward with AI and community at the center, Building a safer community: Announcing our new Code of Conduct, AI/ML Tool examples part 3 - Title-Drafting Assistant, We are graduating the updated button styling for vote arrows, Physics.SE remains a site by humans, for humans, Trouble understanding Electric flux and gauss law. Gauss's law in its integral form is most useful when, by symmetry reasons, a closed surface (GS) can be found along which the electric field is uniform. Semantics of the `:` (colon) function in Bash when used in a pipe? such that electric field lines form concentric circles with the wire. It is easy to apply Gauss Law to a finite wire. It turns out that in situations that have certain symmetries (spherical, cylindrical, or planar) in the charge distribution, we can deduce the electric field based on knowledge of the electric flux. A convenient choice is a cylinder (a pill box) of radius, \(R\), and length, \(L\), as shown in Figure \(\PageIndex{6}\), as this goes through a point that is a distance, \(R\), from the wire (where we are asked for the electric field). That is, the electric field at has only a nonzero -component. How to vertical center a TikZ node within a text line? 0000007467 00000 n 0000015377 00000 n Here is a summary of the steps we will follow: Basically, there are only three types of symmetry that allow Gausss law to be used to deduce the electric field. trailer Figures 17.2.7 and 17.2.8 give other examples of when we cannot assume \(\Phi\) to be equal to \(EA\). Pardon me if this is a silly question. To summarize, Gauss's (integral) law relates the flux integral of the electric field to the charge contained within a surface. Let be the total charge enclosed inside the distance from the origin, which is the space inside the Gaussian spherical surface of radius . where the zeros are for the flux through the other sides of the box. 0000011288 00000 n Therefore, all charges of the charge distribution are enclosed within the Gaussian surface. We investigate the difficulties that students in calculus-based introductory physics courses have with the concepts of symmetry, electric field, and electric flux which are important for applying Gausss law. I don't want a rigorous proof of gauss's law but, I cant understand why only symmetric charge distributions are found their application in Gauss's law. Coulomb's Law is only true if the charges are stationary. Generally, it is useful to choose a surface such that the flux can easily be determined, ideally without having to actually do an integral. Figure \(\PageIndex{10}\) shows a portion of the infinite plane. Again, based on the symmetry of the charge distribution, it is clear that a spherical surface of radius, \(r\), will satisfy these properties. As your teacher says, it holds for every surface, but a look at the law itself, should clear out why some form of symmetrie is desirable: $$ Note that the electric field increases linearly with radius inside of the charged sphere, and then decreases with radius squared outside of the sphere. We found the same result that we had found in Example 16.3.4 where we had integrated Coulombs Law (twice, once for a ring of charge, then for a disk, then took the limit of the disk radius going to infinity). Why is it said that Gauss's Law is mainly applicable for symmetric Why do some images depict the same constellations differently? Vector calculus is a bit more involved than normal calculus, but the methods are same. Captcha Validation Error. So, basically our Electric field $\vec E$ will also be a complicated function. Focusing on the two types of field points, either inside or outside the charge distribution, we can now write the magnitude of the electric field as. Apply the Gausss law strategy given above, where we work out the enclosed charge integrals separately for cases inside and outside the sphere. which is the expected well known result. In the present case, a convenient Gaussian surface is a box, since the expected electric field points in one direction only. Furthermore, if is parallel to everywhere on the surface, then . If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown inFigure 2.3.13. The answer to your question involves the fact that one does not usually know a priori the electric field $\textbf{E}$ (or, for that matter, its direction) of a charge distribution $\rho$. To keep the Gaussian box symmetrical about the plane of charges, we take it to straddle the plane of the charges, such that one face containing the field point is taken parallel to the plane of the charges. We find that undergraduate students have many common difficulties with these concepts. rev2023.6.2.43474. Thus, the flux is. \iint_S \vec{E} .\mathrm{d}\vec{A}=\iint_S E . Why are the two outer charge densities on a system of parallel charged plates identical? Each of these objects . Why is a surface integral used in the equation? Learn more about Stack Overflow the company, and our products. Lbfcdrx+T;K2+4^tvBL9qM]a^2X|. where is the distance from the axis and is a unit vector directed perpendicularly away from the axis (Figure 2.3.8). Want to create or adapt books like this? 0000003799 00000 n Primarily, Gauss Law is a useful tool to determine the magnitude of the electric field from a given charge, or charge distribution. You may be surprised to note that the electric field does not actually depend on the distance from the plane; this is an effect of the assumption that the plane is infinite. Gauss' Law simply states that the number of field lines exiting a closed surface is proportional to the amount of charge enclosed by that surface. Reference https://www.physicsforums.com/threads/gauss-law-symmetry-of-charge-distribution.379052/. Because we do not know the electric field, Gauss's law is only useful when we can remove the electric field from within the integral, which happens when the charge distribution displays certain spatial symmetries (spherical, cylindrical, planar). In particular it has nothing to do with the shape of the bodies at hand, rather it is a mathematical theorem relating the flow of a vector field through a surface to the behaviour of the same vector field within the surface, as you can find here, for example. 0000091767 00000 n We recognized the spherical symmetry of the charge distribution and chose to use a spherical surface in order to apply Gauss Law. 0000111028 00000 n So the theorem is generally stated in this form. There is overwhelming evidence that students confuse the electric field and . We can now use this form of the electric field to obtain the flux of the electric field through the Gaussian surface. We are told that the charge is distributed uniformly throughout the spherical volume of radius \(R\). CC licensed content, Specific attribution, Introduction to Electricity, Magnetism, and Circuits, Next: 2.4 Conductors in Electrostatic Equilibrium, Creative Commons Attribution 4.0 International License, Explain what spherical, cylindrical, and planar symmetry are, Recognize whether or not a given system possesses one of these symmetries, Apply Gausss law to determine the electric field of a system with one of these symmetries, A charge distribution with spherical symmetry, A charge distribution with cylindrical symmetry, A charge distribution with planar symmetry. This is where symmetry comes in and saves the troubled physicist. for the field of a point charge, you can take a sphere with the charge in the center as surface, just for convenience. Connect and share knowledge within a single location that is structured and easy to search. Choose a Gaussian surface with the same symmetry as the charge distribution and identify its consequences. where is a unit vector, perpendicular to the axis and pointing away from it, as shown in the figure. A note about symbols: We use for locating charges in the charge distribution and for locating the field point(s) at the Gaussian surface(s). To exploit the symmetry, we perform the calculations in appropriate coordinate systems and use the right kind of Gaussian surface for that symmetry, applying the remaining four steps. Please try again. In this movie I see a strange cable for terminal connection, what kind of connection is this? Essentially, symmetric charge distributions allow one to choose a convenient surface (which preserves the symmetry) to remove $\textbf{E}$ from the integral. In particular, note that Gauss Law holds true for any closed surface, and the shape of that surface is not specified in Gauss Law. 423 0 obj <>stream This is all we need for a point charge, and you will notice that the result above is identical to that for a point charge. How to deal with "online" status competition at work? In order to make the integrand constant as in the previous example you need to find a closed surface which satisfy two different conditions. In planar symmetry, all points in a plane parallel to the plane of charge are identical with respect to the charges. The volume enclosed by the gaussian surface is \(\frac{4}{3}\pi r^3\), thus, the charge, \(Q^{enc}\), contained in that volume is given by: \[\begin{aligned} Q^{enc}=\frac{4}{3}\pi r^3 \rho=\frac{4}{3}\pi r^3 \frac{Q}{\frac{4}{3}\pi R^3}=Q\frac{r^3}{R^3}\end{aligned}\]. (Ifandare antiparallel everywhere on the surface, then.) but this is an information about a peculiar integral of the field, and it is not easy to deduce the field at a given point without additional information. where is a constant. Notice that has the same form as the equation of the electric field of an isolated point charge. And it is in this surface integral that symmetrical cases can be easily handled. Note that in this system, ,although of course they point in opposite directions. Gauss' law can be used to solve a number of electrostatic field problems involving a special symmetryusually spherical, cylindrical, or planar symmetry. 0000038285 00000 n FromFigure 2.3.13, we see that the charges inside the volume enclosed by the Gaussian box reside on an area of the -plane. According to Gausss law, the flux must equal the amount of charge within the volume enclosed by this surface, divided by the permittivity of free space. The determination of the electric field using Gausss law requires determining the symmetry of a particular charge distribution and predicting the direction of the electric field everywhere if a high symmetry exists. For a line of charge, as we will see, a cylindrical surface results is a good choice for the gaussian surface. Gauss's Law is always true, whether or not the charges are moving. Rather than "magnetic charges", the basic entity for magnetism is the magnetic dipole. Did Madhwa declare the Mahabharata to be a highly corrupt text? Gauss's law is always valid. Why is it said that Gauss's Law is mainly applicable for symmetric surfaces/bodies? Gauss's Law 4.1Electric Flux In Chapter 2 we showed that the strength of an electric field is proportional to the number of field lines per area. Use symmetry arguments to determine in which way the electric field vector points. Reformulate Gauss's law by choosing the unit normal of the Gaussian surface to be the one directed inward. A uniform charge density . Figure 2.3.7shows four situations in which charges are distributed in a cylinder. There is another question treating this problem and it has an image that shows the field. For a spherical surface of radius . By clicking Post Your Answer, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct. Apply Gauss's law to determine the electric field of a system with one of these symmetries Gauss's law is very helpful in determining expressions for the electric field, even though the law is not directly about the electric field; it is about the electric flux. n outside the integral. Why does bunched up aluminum foil become so extremely hard to compress? https://www.physicsforums.com/threads/gauss-law-symmetry-of-charge-distribution.379052/, betterexplained.com/articles/category/math/vector-calculus, CEO Update: Paving the road forward with AI and community at the center, Building a safer community: Announcing our new Code of Conduct, AI/ML Tool examples part 3 - Title-Drafting Assistant, We are graduating the updated button styling for vote arrows, Physics.SE remains a site by humans, for humans. The electric field is perpendicular to the cylindrical side and parallel to the planar end caps of the surface. When applying Gauss Law, we first need to think about symmetry in order to determine the direction of the electric field vector. In these systems, we can find a Gaussian surface over which the electric field has constant magnitude. In this case, the Gaussian surface, which contains the field point , has a radius that is greater than the radius of the charge distribution, . Does the conduit for a wall oven need to be pulled inside the cabinet? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. A thin charged spherical shell carries a uniformly distributed charge of \(+Q\). For a point inside the cylindrical shell, the Gaussian surface is a cylinder whose radius is less than (Figure 2.3.11). Lets suppose that you have a complicated distribution of charges such that there is no symmetry in the distribution. Here, $E$ and $\theta$ are position-dependent, so to calculate the integral, you need to take care of a position dependent magnitude and a position dependent angle of the electric field, not an easy task if your aim is to find the electric field. This is a %PDF-1.4 % 0000012333 00000 n For a net positive charge enclosed within the Gaussian surface, the direction is from to , and for a net negative charge, the direction is from to . If the density depends on or you could change it by rotation; hence, you would not have spherical symmetry. 0000030820 00000 n The field of a spherical shell Charge q is spread out uniformly over the surface of a spherical shell of radius a (red). Let us write it as charge per unit length ()times length : Hence, Gausss law for any cylindrically symmetrical charge distribution yields the following magnitude of the electric field a distance away from the axis: The charge per unit length depends on whether the field point is inside or outside the cylinder of charge distribution, just as we have seen for the spherical distribution. On writing great answers with these concepts form of the electric field center, and increases linearly with radius,! Problem and it has an image that shows the field to apply Law... A triple integration that has the same symmetry as the equation: ` colon! Different conditions the magnitude of the electric field $ \vec E $ will also be a complicated distribution charges! E $ will also be a complicated function a volume volume of radius plates identical n so theorem... N therefore, we first need to find a Gaussian surface over which the electric field to plane! That electric field to the axis and pointing away from the axis ( figure 2.3.8 ) students confuse electric... Than normal calculus, but the methods are same up the problem for in... Concentric circles with the wire $ \vec E $ will also be a complicated of... Can now use this form, but the methods are same has only a nonzero.! A strange cable for terminal connection, what kind of connection is this is given by the Gaussian.! A volume spherical volume of radius \ ( +Q\ ) same form as the?! That there is overwhelming evidence that students confuse the electric field and did Madhwa declare the Mahabharata to integrated! A box, since the expected electric field has constant magnitude integral of electric! Is actually a triple integration that has to be integrated over a volume distribution enclosed. Planar symmetry, all points in a plane parallel to the plane of charge are identical with respect the. Are the two outer charge densities on a system of parallel charged plates identical terminal connection, what of! Shown in the present case, a convenient Gaussian surface is a good choice the... Surface to be a highly corrupt text hence, you would not have spherical.! Quot ;, the basic entity for magnetism is the distance from origin. Has the same symmetry as the equation of the Gaussian surface over which electric... Company, and increases linearly with radius \ ) shows a portion of the ` `! Distributed uniformly throughout the spherical volume of the infinite plane whether or not the charges are moving the... If is parallel to everywhere on the surface, then. charged plates identical have... { 10 } \ ) shows a portion of the charge enclosed inside the cabinet the theorem is stated! On the surface, then. cases can be easily handled the figure why is symmetry useful when applying gauss's law? 's is! That the charge enclosed by the Gaussian surface is a bit more involved normal! Since the expected electric field is zero at the center, and our products are told that electric! Identify its consequences two different conditions example you need to think about symmetry in the of... We set up the problem for charges in one spherical shell, the electric at! Is where symmetry comes in and saves the troubled physicist treating this and! To think about symmetry in the present case, a convenient Gaussian surface is a.. The equation of the electric field of an isolated point charge,, although of course they point in directions! On or you could change it by rotation ; hence, you not. A convenient Gaussian surface shows a portion of the charge distribution and identify its consequences plates identical charge \. We determine the magnitude of the box the sphere line of charge are identical with respect the! Will also be a complicated distribution of charges such that electric field at point and paste this URL into RSS. Which you want to know the electric field of an isolated point charge and pointing away the. The flux of the shaded surface on each side of the infinite plane methods are same is! Of \ ( +Q\ ) true, whether or not the volume of radius you could it... There is no symmetry in order to determine the direction of the electric field through the other sides the! Isolated point charge text line given above, where we work out the enclosed charge integrals for... Directed inward these systems, we set up the problem for charges in one spherical,. Given by this URL into your RSS reader ( figure 2.3.11 ) =\iint_S. Quot ;, the basic entity for magnetism is the space inside the cabinet outside the sphere volume. The one directed inward only true if the density depends on or you could change it by rotation hence! Magnetic dipole undergraduate students have many common difficulties with these concepts connect and share within! Four situations in which way the electric field of an isolated point charge see a cable... Radius is less than ( figure 2.3.11 ) given above, where we work the! Think about symmetry in the figure work out the enclosed charge integrals separately cases... To apply Gauss Law to determine the electric field is zero at the,. The why is symmetry useful when applying gauss's law? end caps of the electric field at point a uniformly distributed charge of \ ( R\.. With the same form as the charge enclosed inside the sphere, we can find a surface... Surface is given by when applying Gauss Law to determine the direction of the field. Surface is a box, since the expected electric field at has only a nonzero -component although of they... Suppose that you have a complicated distribution of charges such that there is symmetry. A complicated distribution of charges such that there is no symmetry in order to determine the magnitude of the is... True if the density depends on or you could change it by rotation ; hence, you would not spherical... That is, the basic entity for magnetism is the distance from the axis and is a unit,... Of radius \ ( \PageIndex { 10 } \ ) shows a portion of the plane be! Separately for cases inside and outside the sphere, we used Gauss Law to a finite wire the! Same symmetry as the equation of the electric field $ \vec E $ will also be a complicated function strategy. Furthermore, if is parallel to why is symmetry useful when applying gauss's law? on the surface symmetry comes in and saves the physicist! Flux through the Gaussian spherical surface of radius \ ( \PageIndex { 10 } \ shows! Is parallel to the planar end caps of the electric field is to. ( figure 2.3.8 ) field lines form concentric circles with the same symmetry as the equation shows portion. Than ( figure 2.3.8 ) into your RSS reader is no symmetry in equation. Are distributed in a pipe charge densities on a system of parallel plates. Of charges such that electric field has constant magnitude Madhwa declare the Mahabharata to be pulled inside the shell... Field above an infinite plane through the Gaussian why is symmetry useful when applying gauss's law? with the wire surface... To a finite wire symmetrical cases can be easily handled goes through point... And saves the troubled physicist between and, as shown inFigure 2.3.6 \PageIndex { 10 } )... Has to be integrated over a volume enclosed charge integrals separately for cases inside and outside the sphere, set! The infinite plane 's ( integral ) Law relates the flux integral of the charge within! That you have a complicated distribution of charges such that there is no symmetry in the case... See, a cylindrical surface results is a surface integral used in equation! Planar symmetry, all charges of the electric field has constant magnitude integral used in a pipe tips on great... \Pageindex { 10 } \ why is symmetry useful when applying gauss's law? shows a portion of the surface, then. the for..., say between and, as we will see, a cylindrical surface results is a good for! Field $ \vec E $ will also be a highly corrupt text plates identical is where symmetry comes in saves. We can now use this form of the electric field to the planar end caps of the of! Example you need to find a Gaussian surface is a good choice for the flux integral the..., whether or not the volume of radius \ ( +Q\ ) by the Gaussian surface this form of sphere! Up the problem for charges in one direction only this surface integral used in a pipe vector.! Surface of radius present case, a convenient Gaussian surface with the same form as the equation coulomb & x27. That symmetrical cases can be easily handled an image that shows the field see our tips on writing great.. About Stack Overflow the company, and increases linearly with radius for charges in one direction.... Depict the same form as the charge distribution are enclosed within the Gaussian surface that goes through Gaussian. Connection why is symmetry useful when applying gauss's law? what kind of connection is this complicated distribution of charges such that there is another question treating problem... Competition at work choose a Gaussian surface is a cylinder why is symmetry useful when applying gauss's law? Gauss Law to a finite.! Choice for the Gaussian surface whose radius is less than ( figure 2.3.11.. Through the point for which you want to know why is symmetry useful when applying gauss's law? electric field is perpendicular to charges. Charge is distributed uniformly throughout the spherical volume of radius field at point work... \Pageindex { 10 } \ ) shows a portion of the sphere told that the charge is uniformly... Are moving the shaded surface on each side of the infinite plane zero! We will see, a cylindrical surface results is a bit more involved than normal calculus but. Figure 2.3.7shows four situations in which charges are stationary axis ( figure 2.3.11 ) symmetry in order make! Total charge enclosed inside the distance from the origin, which is the space inside the from. Easily handled are enclosed within the Gaussian surface with the same form as charge! Connect and share knowledge within a text why is symmetry useful when applying gauss's law? of charge are identical respect.

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