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## Gauss’ Law

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**Gauss’ Law**Besides adding up the electric fields due to all the individual electric charges, we can use something called Gauss’ Law. Its idea is similar to mass conservation in fluid flow involving sources and sinks. To see the idea behind this law and how it works, let’s look at the flow of water.**Gauss’ Law for velocity**Consider the flow of water from a pipe. The amount of water can be measured by the volume, V, per time, t: V/t is a measure of the flow. Volume can be found by multiplying the area by the perpendicular distance: dV = A*ds where ds is always perpendicular to A. We can write this in terms of a dot product: dV = Ads (where the vector A has a direction perpendicular to the surface) so we have dV/dt = A ds / dt = v dA.**Gauss’ Law for velocity**If we integrate over a complete closed surface, then we have dV/dt (created or destroyed) = v dA . However, we know mass (and water) can’t be created or destroyed, so v dA = 0. Electric field is a vector just like velocity is a vector. The sources of electric field are charges. Thus, in analogy with water, we suspect that E dA should be proportional to sources or sinks of electric field. The sources and sinks of electric field we can identify as charges!**Single Point Charge**Consider a point charge. It is the source of an electric field that goes radially out from the point. The Electric Field is not absorbed by space; instead it can only come out of a positive charge or end on a negative charge! This is similar to water - it is not absorbed by space but can only come out of a source or go down a drain.**Single Point Charge**If we enclose this point charge inside a closed area, the amount of electric field coming out of the area will depend only on the amount of charge inside the area and not on the area. The field coming out of the green sphere is equal to the field coming out of the blue sphere and equal to the field coming out of the purple box!**Gauss’ Law**This is just like the flow of water. The flow of water depends on how much comes out of the source (and how much goes into a drain or sink). The mathematical statement of this for Electric fields (Gauss’ Law) is: òòclosed area E · dA = constant * Qenclosed .**Gauss’ Law**òòclosed area E · dA = constant * Qenclosed To find the constant, we look at our point charge case, and we use a sphere for our closed area. In this case, the Electric field for the point charge is: E = kQ/r2, and it points directly away from the charge. The area for each part of the sphere also points directly away from the center of the sphere. Hence, the factor of cos(θ) in the dot product gives cos(0o) = 1 everywhere.**Gauss’ Law**òòclosed area E · dA = constant * Qenclosed Also, since the sphere is equidistant from the center (where the charge is), the Electric field due to the point charge is constant (E = kQ/r2). Hence the integral can be integrated to give: EA = constant*Qenclosedor with the area of a sphere being A = 4pr2 : (kQ/r2)(1)(4pr2) = constant*Q , or: constant = 4pk.**Gauss’ Law**Thus the final statement for Gauss’ Law is: òòclosed area E·dA = 4pkQenclosed . In determining the final form for this relation we put in the known form for the electric field, E. However, we often use this equation to solve for E. But in order to do this, we need to integrate some function that we don’t know. We do this by employing symmetry. (We used the spherical symmetry of a point for the point charge.)**Gauss’ Law and Coulomb’s Law**Actually, Gauss’ Law and Coulomb’s Law say the same thing and are really different expressions of the same law. Recall that a law is something not derived from previous ideas, but rather something that we find by experimentation that works. Sometimes we express the constant k in the form: k = 1 / 4peo , where the sub naught indicates we are working in vacuum.**Hollow Sphere**òòclosed area E·dA = 4pkQenclosed = Qencl/eo . Let’s apply Gauss’ Law to a hollow charged sphere. First, what is the electric field (E) inside a hollow charged sphere? To be definite, consider the point marked with an x in the figure below. By symmetry, Ey= 0 . But what about Ex ? +Q R x**Inside Hollow Sphere**òòclosed area E·dA = 4pkQenclosed . All the charge to the right of the green dotted line causes an electric field to point to the left (away from the charge), while all the charge to the left of the line causes the electric field to point to the right (again, away from the charge). There is less charge to the right, but those charges are closer; there is more charge to the left but they are farther away! Which wins? +Q x**Inside Hollow Sphere**òòclosed area E·dA = 4pkQenclosed . Let’s try using Gauss’ Law. We first need to choose an area. To take advantage of the symmetry, let’s choose a sphere that goes through our point (blue sphere). By symmetry the integral reduces to: EAblue sphere = 4pkQenclosed in blue sphere . However, Qenclosed = 0, and so Einside = 0 . +Q x**Outside Hollow Sphere**Now we’ll consider the electric field at a point outside the charged sphere. We’ll again choose a definite point marked with the blue x on the diagram below. To use Gauss’ Law, we need to choose a definite closed area, which, again using the symmetry, will be a sphere going through our point, x. +Q R x r**Outside Hollow Sphere**òòclosed area E·dA = 4pkQenclosed . Again, by symmetry the integral can be integrated to give: EA = 4pkQenclosed . Thus we have, with A = 4pr2 , E =4pkQenclosed / 4pr2 = kQ/r2 . This is just the same formula we had for a point charge! +Q R x r**Charged Sphere**Hollow Sphere Results using Gauss’ Law: Inside, E = 0 . Outside, magnitude: E = kQ/r2 and direction points away from the center of the sphere(just like a point charge). What about the case of a SOLID sphere of charge? (In this case, the charges must be locked into place, because they would normally try to get as far away from each other as possible - and end up on the surface - the same situation as a hollow sphere of charge!)**Solid Sphere of Charge**We can consider that a solid sphere of charge is made up of a bunch of hollow spheres of charge that have a common center. Thus the charges on the outside of any sphere do not contribute to the electric field, whereas the charges inside a particular hollow sphere would contribute to the field at the hollow sphere.**Solid sphere of mass**Do gravitational fields act just like electric fields? YES, only the charges are replaced by masses and the k is replaced by the G. Is there a Gauss’ Law for gravitational fields? YES! òòclosed area g·dA = 4pGMenclosed What about the gravitational field due to a solid sphere? We have just shown that the gravitational field should be exactly like a point mass at the center of the sphere. (This is what we assumed when we considered the gravity of the earth in PHYS 150 !)**Gauss’ Law: other symmetries**What other symmetries might we be able to use Gauss’ Law for? • We’ve already considered spherical symmetry which works for points and spheres. • How about cylindrical symmetry which should work for charged lines and cylinders? • How about plane symmetry which should work for plates of charge.**Line of Charge**Consider a long line of charge that has a uniform charge density, l = Q/L . What is the electric field strength at a point (marked x) a distance a away from the wire? By considering symmetry, we know that the Electric Field,E,must point directly away from the wire. (We already did this problem the straightforward way, but let’s do it using Gauss’ Law for practice.) E x a l=Q/L**Line of Charge**òòclosed area E·dA = 4pkQenclosed . To use Gauss’ Law, we need to choose an area based on the symmetry of the situation. Let’s choose a cylinder of length L and radiusathat goes through our point. The cylinder has three separate areas: the left side, the right side, and the “can” part. x a l L**Line of Charge**For a long wire, we know by symmetry that the Electric field must point away from the wire and so must be perpendicular to the area on both the left and right sides. Because of the dot product between Eand dA, these two parts of the cylinder do not contribute to the integral. x E E dA a dA l L**Line of Charge**òòclosed area E·dA = 4pkQenclosed . For the “can” part, however, E and dA are parallel, and so the cos(θ) factor in the dot product in Gauss’ Law gives 1. Also, by symmetry, the magnitude of the electric field is constant along the can part. Thus Gauss’ Law gives: EAcan = 4pkQenclosed . With Acan = 2paL and Qenclosed = lL, we have E = 4pk lL / 2paL = 2kl/a . dA E x a l L**Line of Charge**E = 2kl/a directed away from the positively charged wire. Note that this is the same formula for electric field that we obtained using the straightforward way. However, the straightforward way involved a hard integral, whereas the Gauss’ Law way involved a simple integral (once we chose the right area based on the symmetry).**Cylinder of Charge**If we have a long cylinder of charge, we can again use Gauss’ Law to determine the electric field inside and outside the cylinder. l = DQ/DL**Insidea Cylinder of Charge**For a point inside the cylinder, we can choose an area in the shape of a cylinder that goes through our chosen point, x. Because on the left and right sidesEand dA are ^, they do not contribute. And by symmetry, the “can” part is constant to give EAcan = 4pkQencl. But since Qencl = 0, Einside= 0 ! l = DQ/DL E dA x E R dA**Outside a Cylinder of Charge**For a point outside the cylinder, we can choose an area in the shape of a cylinder that goes through our chosen point, x. The left and right ends of the cylinder do not contribute since E and dA are perpendicular. E x dA l = DQ/DL**Outside a Cylinder of Charge**The “can” part has E and dA parallel, and the E is constant along the “can”. Hence Gauss’ Law gives: EAcan = 4pkQenclosed . With Acan = 2paL and Qenclosed = lL, we have Eoutside = 4pk lL / 2paL = 2kl/a , the same as for a long line of charge! dA E x l = DQ/DL**Charge on Plates**We have considered a point (spherical) and a line (cylindrical). Now we extend our analysis to a plane (plate) of charge. We are looking for the E field at a point, x, at adistance a above a large plate of charge with charge density s = Q/A. Again by symmetry, we know the E field points away from the plate. E x a s=Q/A**Above a Charged Plate**òòclosed area E·dA = 4pkQenclosed . To use Gauss’ Law, we need to choose an area based on the symmetry of the situation. Let’s choose a box of length L that has a side going through our point (so a = L/2). E x a=L/2 s=Q/A L L**Above a Charged Plate**òòclosed area E·dA = 4pkQenclosed . The four sides have dA’s that are horizontal, while the E field is vertical. Thus the dot product in Gauss’ Law says these do not contribute. However, both the top and bottom sides have areas that are parallel to the field. Hence we have EAtop + EAbottom = 2EA = 4pkQencl, or E = 2pks. E dA x a=L/2 s=Q/A dA dA L L**Review**For a point or outside a sphere, E = kQ/r2. For a line or outside a cylinder, E = 2kl/a . For a plate of charge, E = 2pks = s/2eo . There are several things to note: 1. All the formulas have a k in them. • a) The field due to a point falls off as r-2, b) the field due to a line falls off as a-1, c) whereas the field due to a plate is constant! How come? And what about units?**Units**As to units, note that the line and cylinder formula has a l in it, and l has an inverse distance in it (l=Q/L) to go with the a-1. The plate formula has a sin it, and recall s=Q/A, and A has distance squared in it. So the units do work out!**Physical Behavior**But why the difference in strength with distance for the three cases? Note that, for the line of charge, as the field point gets farther away from the line, the charges further along the line do not get that much farther away (green line versus blue line) but the angle gets steeper, so the vertical component actually gets bigger. This somewhat larger vertical component somewhat diminishes the lesser effect the charges right underneath the point have on the field as the point gets farther away. x x**Physical Behavior**The same thing happens for the plate, except the plate is in two dimensions instead of just one. In other words, at any particular distance from the point on the line there are only two charges (one on the left and one on the right); for the plate there is a whole ring at any particular distance, and as the distance gets bigger, the size of the ring, and hence the amount of charge on the ring, gets bigger! This increase in charge and increase in vertical component of the further charges cancels the decrease in strength of the nearer charges exactly.**Special Case: Coaxial Cables**What is the electric field around a coaxial cable that has a positive charge on the (inner) wire and an equal but opposite charge on the (outer) cylinder? All we have to do is add the fields due to each. (I leave it to you to investigate this. What is the electric field between the inner wire and the cylinder? What is it outside the cylinder?)**Special Case: Parallel Plates**Consider the situation of having two plates that are parallel with one plate having a positive charge and the other plate having an equal but opposite charge. What is the electric field outside the two plates? What is the electric field between the two plates?**Computer Homework for Gauss’ Law**The 2nd computer homework program on Vol. 3 is on Gauss’ Law and will give you practice in applying the results to specific problems.