Chapter 24: Gauss Law

Transcription

1 Chapter 4: Gauss Law In Chapter, the tool symmetry was very important in simplifying problems. Gauss law takes these symmetry arguments and maximizes their efficiy in simplifying -field calculations. Coulomb s Law Method is barbaric and uses pure brute force when solving for the -field. We derived some equations using some fairly strenuous integration. This is the hard way of doing things. Coulomb's method for solving electrostatics can be formidable - much of the rest of electrostatics is devoted to assembling a bag of tools and tricks for avoiding these integrals There is a saying that goes something like this: there is a hard way and easy way to do the same job the easy way usually involves using the right tools. Gauss law is the right tool. Gauss Law Method is a more elegant approach for solving -fields when there is high degree of symmetry. Consequently, we will be restricted to a limited number of charge distributions. As you will see, we will be able to derive some -fields in only a few lines! The basic idea of Gauss' law is. nclose a charge distribution with an imaginary or Gaussian surface. We then mathematically count the number of -field lines poking through this Gaussian surface. This counting is the electric flux.. If this Gaussian has the right symmetry, we will then be able to relate the Gaussian surface area to the -field function. This may sound rather like an indirect method of expressing things, but it turns out to be a tremendously useful relationship. What is a Gaussian A Gaussian is an imaginargy closed surface with a high degree of symmetry. If it has the proper symmetry, then a Gaussian is a constant -field surface. The key point being that if you cannot do this, then it is easier to use Coulomb s Method for determining the -field. Gaussian Surface constant -field surface There are only types of Gaussians : spherical, cylindrical, and cubic ( pill box ) shell. The connection between lectric Flux and lectric Charge Definition: The number of -field lines poking out or in through the Gaussian is measured by the electric flux. lectric Flux Φ = da = number of -field lines passing through a Gaussian An losed charge creates a flux through a Gaussian tells us what is going on inside. In other words, we will be able to determine the magnitude and sign of the losed charge. Here are three key points about Gaussians and flux: i. The electric flux is directly proportional to the losed charge. ii. Charges outside a Gaussian will not contribute to the flux of the losed charge. iii. Flux is independent of size and shape of the Gaussian. Part : lectric Flux is proportional to the nclosed lectric Charge (Φ q) Suppose an electric charge lines within the Gaussian, (i) how much and what sign does the lose charge have? 4.

2 HW applet Gauss law Suppose a positive charge is losed by a spherical Gaussian. The electric charge will produce an electric flux that pokes through the Gaussian. Measuring the amount of flux through the Gaussian will tell us the number of field lines coming through the Gaussian. That is, if four lines poking through is for one electric charge, then 8 lines is twice the charge losed. If the flux is pointing outwards, then the losed charge is positively charged whereas inwards, its negatively charged In summary, ( lectric Flux) ( nclosed lectric Charge) Φ ± q ± q losed ii. lectric charges outside a surface will not contribute to the electric flux of q Suppose a Gaussian loses some electric charges AND there are different, external electric charges outside the Gaussian. The net flux produced by these external electric charges not affect the flux reading of the losed charge. q = q Φ net = Φ q +Φ + +Φ =Φq cancel out iii. The lectric Flux is Independent of the Size and Shape of the Gaussian If the losed charge is fixed, then the flux is q = Φ = = A = constant constant A Consider two spherical Gaussians where one of the s has twice the radius as the other. Since the magnitude of the field decreases as /r, the field at the surface of a twice as large has its -field reduced by ¼ when compared to the other : vs. = r (r) 4 r On the other hand, a is twice as large, has four times the area: A = 4π r vs. A = 4 π (r) = 4 4π r Comparing the fluxes through both s tells us Φ = A and Φ = A = 4 4A Φ = Φ The lectric Flux is Independent of the size of the Calculating the lectric Flux The electric flux depends on the -field and the area. The area vector is a vector that is normal to the surface: A = An = (A, A ). It is customary to pick the outward direction to be the positive normal direction. There are area vector components: 4.

3 Let s first start by viewing this applet. Applet Davidson s Flux The amount of flux (i.e., number of field lines poking through the losed area) is the dot product between and da. Flux Φ = da = cosθda where Units: [ ] [ ][ ] + out of the surface A = An ˆ nˆ = normal = into the surface N/C N m /C Φ = A = m = However, since the Gaussian is independent of area and not going into the details of the situation, when the dust settles, the important detail for now is the cosine angle between and da. The angle dependent of the flux shows that there are three possible angles but only two will be important for our purposes: θ = and 9. Question: A rubber balloon has a single point charge in its interior. Does the electric flux through the balloon depend on whether or not it is fully inflated? xplain your reasoning. Gauss Law (Stated without Proof) The flux through any kind of surface is equal to the losed charge divided by ϵ. The -field is the total electric field at each point on the Gaussian Surface. A Deeper look at the vector - field From vector calculus, there is a mathematical theorem known as Gauss law that relates an electric field vector on a surface area to the diverge in a volume: da = d where the diverge is number of field lines coming through a Gaussian surface Here is a much deeper inside view of what an electric field is and its physical meaning. An ordinary vector can be multiplied different ways: aa, A B, A B The same applies to three-dimensional derivatives or Del Operators: A, A, A gradient diverge curl Gauss' Law: q Φ = da = surface 4.

4 What do these mathematical operations measure?. Gradient is the standard derivative which measures slope. The diverge mathematically looks like A = ( x, y, z) ( A x,a y,az) = xax + yay + zaz Geometrical interpretation: A is measure of how much the vector A spreads out (diverges) from a point in question.. Will save this one for when we get to magnetism but here is a little bite : The curl A is measure of how much the vector A curls (rotates) about a point. xample 4. A charged particle is held at the center of two concentric conducting spherical shells. The figure on the right gives the net flux Φ through a Gaussian centered on the particle, as a function of the radius r of the. What are (a) the charge of the central particle and the net charges of (b) shell A and (c) shell B? Solution The point charge at the center has no dimensions but the shell A has a width being r r and while shell B is r 4 r. a. To determine q inside of shell A, a Gaussian (of radius r a) is placed inside r (r a < r ). From the plot, we read that Φ = 9. 5 in SI units. Substituting this into Gauss law, we get 8.85 C / N m 5 ( )( ) Φ= q / ε q =εφ= 9. N m / C = 8µ C = q b. To determine q outside of shell A, a Gaussian (of radius r b) is placed between r > r b > r. The value of the flux is +4. 5, which implies that q (r < r ) central 5 ( )( ) b =εφ= 8.85 C / N m 4. N m / C =.5 µ C But we have already accounted for some of that charge in part (a), so the result is q (r < r ) = q + q q = q (r < r ) q b A central A b central =.5 µ C ( 8µ C) =.5 µ C µ C = q This makes physical sense because to go from a large negative to a positive flux requires a lot of positive charge. c. To determine q outside of shell B, a Gaussian (of radius r c) is placed outside of r 4 (r c > r 4). The flux value is. 5, which implies q (r < r ) 5 ( )( ) c 4 =εφ= 8.85 C / N m. N m / C =.77 µ C Putting it all together the result is A 4.4

5 q (r < r ) = q + q + q q = q (r < r ) q q c 4 A B central B c 4 A central =.77µ C µ C ( 8µ C) = 5.7 µ C 6 µ C = q B Applications of Gauss Law (Spoken emarks). Gauss Law is most useful and should only be applied to cases where there is a high degree of symmetry. When there is not a high degree of symmetry, it may be easier to use Coulomb s Method to get the electric field.. Gauss' law is almost of magical power and it all hinges on the fact that Coulomb's law has a /r -character. Without this character, Gauss' law is not possible. Gauss law is most commonly used if the losed electric charge is known, what is the electric field produced by this losed charge?. Once a gaussian has been chosen, the -field function is mathematically pulled out of the integral because it is a constant everywhere on the gaussian surface: q Φ = da = cosθ da = q / = da Step POBLM SOLING STATGIS There is a -step process for applying Gauss law to derive the form of the electric field. Step : Pick a good Gaussian Surface Step Symmetry condition (pulls = constant out of the integral sign) Dot-product condition evaluates the cosine function either as ± or. Step : valuate the losed charge q Step : Apply Gauss law to derive the -field expression Step : Picking a good Gaussian The goal is to pull out of the flux integral the electric field function and the cosine term. A good Gaussian satisfies the (i) Symmetry and (ii) Dot-Product conditions. Symmetry Condition: The symmetry of the charge distribution must also be the symmetry of the Gaussian surface: choose Spherically symmetric charge distribution Spherically Gaussian shell Cylindrical symmetric charge distribution Planar symmetric charge distribution choose Cylindrically Gaussian shell choose Cubically Gaussian shell The job of the symmetry condition is to pull the -field variable out of the flux integral sign: da = cos θ da surface surface That means that the electric field will have the same value for most, if not all, everywhere on the gaussian. Picture wise, we get Dot-Product Condition: The Dot Product between and da yields a cosine function: da = ( da)cosθ. We look at the direction of the area and -field vectors, where are three possibilities: da θ= cos( ) = + positive flux (antiparallel) da θ=8 cos(8 ) = negative flux da θ=9 cos(9 ) = negative flux The result of step is 4.5

6 surface da = cos θ da surface Step Step : valuating the nclosed Charge q There are two possible situations for the charge distribution through the volume: (i) constant or (ii) variable. Constant charge distribution A constant charge distribution is described by its charge distribution function defined by the symbol rho ρ. If rho is constant () then the losed charge is defined via the losed charge equation: Q q ρ = q = Q where [ ρ ] = C/m total total There are two situations that will be ountered in problem sets. Point P outside the charge distribution If the Gaussian loses the total charge, then the losed charge is the total charge: q (point outside the surface) = Q = ρ Point P inside the charge distribution If the Gaussian loses only part of the total charge, then the losed charge is q (point inside the surface) = Step : Applying Gauss Law Substituting Steps & into Gauss Law, solving for the -field function gives da da surface surface GaussiandA surface total total q Φ= = = = One is left with an integral over the area of the Gaussian surface: cylinder cylinder 4 = π r L = πr cylinder A = da = π rl A = da = 4πr Cylinder = Sphere = I will now apply this -step process for several examples; insulated solid, spherical shell, a thin wire (or cylinder), and an infinite plane sheet. Insulating Solid Sphere An insulating solid of radius has a uniform volume charge density ρ and carries a positive total charge of q total = Q. (a) Derive the -field using Gauss s law (show the details) for < r <. (b) Plot (r) vs. r for < r <. Solution a. When ever the charge distribution is said to be uniformly charged, it means that the charges are in the inside as well as on the surface. There are two possibilities: outside and inside the. i. (r > ), outside the solid insulating Step : Pick a good Gaussian Q q 4.6

7 The charge distribution has spherical symmetry, so the Gaussian is a spherical shell. One draws a spherical Gaussian with a dashed line at a distance r that loses the total charge. The flux at the Gaussian surface has and da parallel, so the cos θ function is cos () = +. The flux integral is therefore Φ = da = cosθ A spherical shell = + 4π r =Φ (spherical shell) Step : Determine the losed charge Since the Gaussian loses all of the charge of the solid, the magnitude of the losed charge is q = Q = Q emember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian. Step : Apply Gauss law to determine ( < r < ) Gauss law gives Q da = q / + 4π r = Q/ (r > ) = 4 π r Step Step ii. ( < r < ), inside the solid insulating Step : Pick a good Gaussian The charge distribution inside the insulating still has spherical symmetry, and we have already solved this for a spherical Gaussian. Since it has already been done, we just take the result: Φ = + 4π r Step : Determine the losed charge Is there an -field inside a charge insulator? Absolutely! Would one expect a larger - field inside the charged insulating solid? NO why? If the Gaussian loses less charge, then the flux is smaller and this leads to a weaker -field. Inside an insulating, one must use the losed charge equation to determine q. q qtotal ρ = constant = = q = Q total total There are two different volumes in this equation: 4 4 total() = π πr r 4 = for r 4 = < < (Gaussian) = πr total π That is, the losed charge q is less than the total charge Q since the ratio of volumes is less than one. Solving for the losed charge gives q = Q = Q = q total emember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian. Step : Apply Gauss law to determine ( < r < ) Gauss law gives r 4.7

8 r Q da = q / + 4π r = Q/ ( < r < ) = r π Step In summary, 4 Step insulating Q (inside) r, < r < 4π = Q (outside), r > 4π Interpretation of esults. A plot of (r) versus r is shown on the right. Note that inside is different from outside this is a key signature of -field across boundaries ( in out). What would happen if the field was not discontinuous? The field would continue to behave as /r for r <, so that the field would then be infinite at r =, which is physically impossible. However, Gauss' law shows that in the limit as the radius of the Gaussian goes to zero, there is a smaller and smaller losed charge until there is no losed charge. So, the -field goes to zero: lim r inside This inside result eliminates the problem that would exist at r = if (r) varied as /r inside the as it goes outside the. Side Note: This is a problem with classical electromagnetic theory and modern particle physics. Point particle have infinite fields at their origin. How does one eliminate this type of problem? One extends a point particle into a string and these infinities vanish! Superstrings however, it has its own problems.. These two expressions for the electric field match at the boundary (surface): q q inside = r = = 4πε 4 ε total total π r= outside. For those of you who have had Phys 4C or remember Snell s law from optics, the index of refraction occurs because of the electrical properties of a dielectric. That is, at the boundary of a dielectric, the -fields are not the same just inside and outside a in out dielectric ( ). This is the main reason why light is refracted across the boundary and therefore, why a change in the index of refraction occurs. My conundrum with Gauss law in electrostatics If I use Gauss law to calculate the electric field outside of a charged (conducting or insulating) or a point charge, the fields are the same. However, as a test approaches a point charge, the field becomes infinite whereas the test charge approaching the surface of the reaches a finite value of Q/4πϵ, where and Q are the radius and charge of the. Why wouldn t the field also go to infinity as I get infinitesimally close to the surface of? Those are just point charges on the surface. What ultimately led me to this question was trying to physically understand why Gauss law predicts the same field for a charge point and charged (obviously r >). Answer: there are two ways to see out of this conundrum.. A physical way to see the answer is to ask what the test charge sees as it approaches the surface of the conductor. At first, with a test charge far from the 4.8

9 charged, it sees the as point-like with a field value of Q/4πϵ r, where r is the radius of the Gaussian spherical shell. However, as the test charge gets really close to the surface of a conducting, the surface of the appears as an infinite plane whose field value is Q Q/4π σ (r ) = = = 4π In other words, as the test charge approaches the conductor s surface, it s field-ofview is reduced and sees less field lines. This compensates for the diminishing influe of any particular piece, giving the electric field a fixed and constant value at the surface. In other words, at really small distances from the surface, the conductor is an infinite plane, as shown above.. The "point charge" you approach has an infinitesimal charge (thus cancelling out and giving a net zero field). There are infinite such points, true, but you don't approach all of them just one. The rest combine to give a finite value, Q/4πϵ. Gauss law predicts the same field because the net losed charge is the same and has the same symmetries for r >. You might the follow question: How does the rest of the charges combine to give the finite value of the 's surface? Answer: We do not approach the rest of the charges, so the r for them is not zero. ach one is a finite distance away, and as each one is infinitesimal charge the force by each one is infinitesimal. In this case, the infinite sum of infinitesimals converges. Conducting Spheres Thin Spherical Shell A thin spherical shell of radius has a total electric charge Q distributed uniformly over its surface. (a) Derive the -field using Gauss s law (show the details) for < r <. (b) Plot (r) vs. r for < r <. Solution i. (r > ), outside the conducting spherical shell The calculation for the electric field outside the shell is identical to the previous one for the insulating solid : Q r > =, r > (outside) 4π ( ) ii. ( < r < ), inside the solid insulating Since there is no losed charge, there cannot be any flux produced within the shell, and therefore, no electric field exists within the spherical shell.: q = Φ = in =, r < ( inside) One can obtain the same result using Coulomb s method; however, this type of calculation is rather complicated. Gauss law allows us to determine these results considerably easier no doubt. Conductors in lectrostatic quilibrium Good conductors have charges that are NOT bound to any one atom and therefore, charges are free to move within a material. When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium. A conductor in such a state has the following properties: Basic Properties of Conductors 4.9

10 (i) = inside a conductor. Why? Because if there were any field, those free charges would move, and it wouldn't be electrostatics anymore (If this was not so, picking up a piece of steel (say, a fork) would have a current flowing through it and could possibility shock you if you somehow tap into this current.). Well... that's hardly a satisfactory explanation; maybe all it proves is that you can't have electrostatics when conductors are present. We had better examine what happens when you put a conductor into an external -field. Initially, this will drive any free negative charges to the left, but when they depart the right side is left with a net positive charge (due to the stationary nuclei). When they come to the edge of the material, the charges pile up: plus on the right side, minus on the left. Now, these induced charges produce a field of their own, induced, which, as you can see from the figure, is in the opposite direction to external. That's the crucial point, for it means that the field of the induced charges tends to cancel off the original field. Charge will continue to flow until this cancellation is complete, and the resultant field inside the conductor is precisely zero. The whole process is practically instantaneous ( 6 s). This is called a Faraday Cage a hollow conductor (ii) ρ = q = inside a conductor. This follows from gauss's law ( = ρ/ε ). If =, so also is ρ. There is still charge around, but exactly as much plus charge as minus, so the net charge density in the interior is zero. (iii) If an isolated conductor carries a charge, it resides on its surface. That's the only other place it can be. I think it is strange that the charge on a conductor flows to the surface. Because of their mutual repulsion, the charges naturally spread out as much as possible, but for all of them to go the surface seems like a waste of the interior space. Surely we could do better, from the point of view of making each charge as far as possible from its neighbors, to sprinkle some of them through the volume... Well, it simply is not so. You do best to put all of the charge on the surface, and this is true regardless of the size or shape of the conductor. A better way to phrase this problem is in terms of energy. Like any other free dynamical system, the charge on a conductor will seek the configuration that minimizes its potential energy. What property (iii) asserts is that the electrostatic energy of a solid object (with specified shape and total charge) is a minimum when the charge is spread over the surface. For instance, the energy of a if the charge is uniformly distributed over the surface is lower compared to that of a with charge uniformly distributed throughout the volume. (iv) is perpendicular to the surface, just outside a conductor. Otherwise, as in (i), charge will immediately flow around the surface until it kills off the tangential component. (Perpendicular to the surface, charge cannot flow, of course, since it is confined to the conducting object.) The electric field very close to the surface of a charged conductor is perpendicular to the surface and has a magnitude of surface = η/ϵ. Why? Because all of the field lines are pointing in one direction for a conductor and with the infinite plane, they are pointing both directions. So the conductor must have twice the line density as an infinite plane. 4.

11 (v) On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is smallest, i.e., sharp points (stated this without proof). Applications of a Faraday Cage (lectric shielding) DMO radio in Faraday cage and telephones in aluminum foil Youtube (i) Partial vs. Full Faraday cage, (ii) Tesla coil with cage, (iii) Faraday Suit xamples: MI, cellphones on trains, cars struck by lightning, and Theater of lectricity in Boston xample 4. A solid conducting of radius is concentric with a spherical conducting shell of inner radius and outer. The has a charge q ; the shell has a net charge q = q. (a) What is the net charge on the inner and outer surface of the shell? (b) Derive the -field using Gauss s law (show the details) for < r <. (c) Plot (r) vs. r for < r <. Solution a. The inside has charge q while the spherical shell has charge q. The inside solid must have all of its charge q on its surface (as already discussed). Charge q induces (or pulls in) the oppose charge of q on the inner surface (at ) of the spherical shell. This then leaves charge q on the outer surface (at ). b. The charge distributions have spherical symmetry and therefore, the electric flux calculations are all identical. Let s calculate the fluxes starting from the inside ( < r < ) and then move outwards. iii. ( < r < ), between the solid and inside surface of the shell Step : Pick a good Gaussian The charge distribution of the inside charged has spherical symmetry, so the Gaussian is a spherical shell. Drawing (red circle) a Gaussian of radius r between the (at ) and the inner surface (at ). The flux through the Gaussian has and da parallel, where cos () = +. The flux integral is therefore Φ = da = cosθ A = + 4π r =Φ (spherical shell) spherical shell Step : Determine the losed charge Since the Gaussian loses all of the charge of the solid, the magnitude of the losed charge is q = q = q emember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian. 4.

12 Step : Apply Gauss law to determine ( < r < ) Gauss law gives q da = q / + 4π r = q / ( < r < ) = 4 π r Step Step iv. ( < r < ), inside the shell Step : Pick a good Gaussian To determine the flux for a Gaussian inside the shell ( < r < ), one draws a Gaussian a distance r (which is different from the previous calculation. The charge distribution is still spherically symmetric; therefore, we can use the previous result for the flux through the Gaussian and write da = + 4πr Step : Determine the losed charge Since the Gaussian loses the negative charge of the inside surface of the shell AND the positive charge of the solid, the magnitude of the losed charge is q = q + q = q + q = = q inside solid surface Once again, the charge on the outside surface of the shell will not contribute to the net flux because it is outside this Gaussian. One immediately concludes that ( < r < ) = Step : Apply Gauss law to determine ( < r < ) Gauss law gives q da = q / + 4π r = q / ( < r < ) = 4 π r Step Step v. (r > ), outside the shell Step : Pick a good Gaussian In this case, one draws a Gaussian outside the shell (r > ), where it loses everything. Although the calculation of the flux is the same as in the two previous cases, the cosine is now negative since and da point in opposite directions: cos(8) =. The flux is da = 4πr Step : Determine the losed charge Since the Gaussian loses everything, the magnitude of the losed charge is q = q q + q = q = q emember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian. Step : Apply Gauss law to determine ( < r < ) Gauss law gives q da = q / 4π r = q / (r > ) = 4 π r Step Step 4.

13 In summary, the -fields for this configuration are (r < ) = + q ( < r < ) = (r) = ( < r < ) = q (r > ) = 4πr 4πr xample 4. A thick spherical shell has uniform volume charge density ρ, inner radius and outer radius =. (a) Derive the -field using Gauss s law (show the details) for < r <. (b) Plot (r) vs. r for < r <. (c) Numerically calculate the -field at (i) r = /, (ii) /, and (iii) where =. cm and ρ =.84 nc/m. Solution a. The description of the charge density in the problem statement says that the spherical shell is uniformly charged, which implies it is an insulator. Furthermore, since it is a, the flux part of Gauss law will always be the same, and is given by da = + 4πr The electric field calculation will only depend on the nature of the losed charge that is q (r) = 4 π r The focus now is to determine the losed charge for each of the three regions. i. (r < ), inside of inner surface Since there is no charge inside of, the losed charge and (r < ) are zero: q = (r < ) = ii. ( < r < ), inside the shell To determine the losed charge, we start off with the charge density equation: Qtotal q ρ = constant = = q = Q total total total The lose charge depends on two terms, total and where Q total is fixed. The total volume is calculated by subtracting two volumes: total = ( 4 ) ( ) = π ( ) The same applies to the losed volume but the two radii are r and : = (r) ( 4 ) = π (r ) The losed charge is then 4.

14 q = Qtotal = total 4 ( π r π( ) ( r ) Q = Q = q ) 4 ) Now, we can write the electric field for in-between the shells: ( q r ) ( r ) Q 4 r 4 r ) < < = = π π ( ( Q r ) = = ( < r < ) 4π r ) ( iii. (r > ), outside the shell When the Gaussian is drawn outside the shell, the total losed charge is Q, and so the electric field is Q (r > ) = 4 π r In summary, the -fields for this configuration are (r < ) = Q ( r ) (r) = ( < r < ) = 4πr ( ) (r > ) = ( Infinitely Long, Thin Wire A common type of coaxial long thin wires is (i) current carrying wires and (ii) coaxial wires: xample 4.4 An insulating rod of radius and length L has a uniform volume charge density is inside a thin-walled coaxial conducting cylindrical shell of radius >. The net charge on the rod is 4q and on the shell is +q. (a) Derive the electric field using Gauss s law (show the details) for < r <. (b) Plot (r) vs. r for < r <. Solution How are the charges distributed on the cylinder? If the insulating rod has a net charge of 4q and the thin-walled conducting cylindrical shell is +q, the net charge outside the coaxial cable is q(r > ) = 4q +q = q whereas q( < r < ) is variable because it is insulating. i. (r > ), outside the cylindrical shell Step : Picking the Gaussian Surface Symmetry Condition: the charge distribution is cylindrically symmetric, so one chooses a cylindrical Gaussian of arbitrary radius r and arbitrary length L. From the previous chapter (), a line charge 4.4

15 has an electric field that is perpendicular to the surface of the wire. A fixed distance r perpendicular to the wire is where the Gaussian is a constant electric field surface. However, there are two area parts to this cylindrical Gaussian, the side walls and the end caps. The electric flux integral is then broken into these two area parts: da Φ =Φ +Φ = + da end pieces curved end pieces curved side walls side walls Dot-Product Condition: looking at the image above, the end pieces have perpendicular to da (cos 9 = ), and therefore, there is NO flux through the end pieces: da = da cosθ = = Φ θ= 9 end pieces The curved side walls have and da parallel to each other (cos 8 = ), so that Φ = Φ = + da = π rl = Φ cylinder side walls side walls cylinder Now remember, all cylinders have this very relationship regardless of their charge density. Step : valuating the nclosed Charge q Since the Gaussian loses the whole wire, the losed charge is q = 4q + q = q = q Step : Applying Gauss Law Applying Gauss Law gives us Φ = da = q / π rl = q/ side walls Step Step q/l λ (r > ) = = = (r > ) infinite infinite cylinder πr πr cylinder ii. ( < r < ), in-between the cylinder and cylindrical shell Step : Picking the Gaussian Surface The charge distribution inside the insulating cylinder still has cylindrical symmetry, and we have already solved this for a cylindrical Gaussian. Since it has already been done, we have Φ = π rl Step : Determine the losed charge Since the Gaussian loses the whole wire, the losed charge is q = 4q = 4q = q Step : Apply Gauss law to determine ( < r < ) Gauss law gives 4λ da = q / π rl = 4q/ ( < r < ) = π r Step Step iii. ( < r < ), outside the cylindrical shell Step : Picking the Gaussian Surface The charge distribution inside the insulating cylinder still has cylindrical symmetry, and we have already solved this for a cylindrical Gaussian. Since it has already been done, we just take the result: 4.5

16 Φ = π rl Step : Determine the losed charge Inside an insulating cylinder, one must use the losed charge equation to determine q. q qtotal ρ = constant = = q = Q total total There are two different volumes in this equation: total(cylinder) = π L πrl r = for r = < < (Gaussian) = πr L total πl That is, the losed charge q is less than the total charge Q since the ratio of volumes is less than one. Solving for the losed charge gives q = Q = Q = q total emember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian. Step : Apply Gauss law to determine ( < r < ) Gauss law gives r 4λ da = q / π rl = (4q/ ) ( < r < ) = r π Step In summary, Step 4λ r, < r < π 4 λ =, < r < 4 λ, r > π r ) concentric cylinders π r r Application: Geiger counter and concentric cylinders DMO Geiger tube A Geiger tube consists of a tube filled with a low-pressure inert gas. The tube contains two electrodes, between which there is a voltage of 4 6. The walls of the tube are either metal or have their inside surface coated with a conductor to form the cathode while the anode is a wire passing up the center of the tube. When ionizing radiation strikes the tube, some of the molecules of the fill gas are ionized, either directly by the incident radiation or indirectly by means of secondary electrons produced in the walls of the tube. This creates positively charged ions and electrons, known as ion pairs, in the fill gas. The strong electric field created by the tube's electrodes accelerates the positive ions towards the cathode and the electrons towards the anode. Close to the anode the electrons gain sufficient energy to ionize additional gas molecules, creating an avalanche which is collected by the anode. This is the gas multiplication effect. 4.6