Chapter 22 - The Electric Field II Continuous Charge Distributions

Chapter 22 - The Electric Field II: Continuous Charge Distributions ● Point charges: ● By analogy for a charged solid, ● Consider the solid as a collection of an infinite number of point charges, ● ● . Charge Density - a charge distributed across an object o Linear Charge Density - o Area Charge Density - o Volume Charge Density - Two major cases for use of Linear Charge Density ( ) IMPORTANT STUFF know these two proofs!!! CASE 1 - Electric field at the center of a CASE 2 - Electric field along the central charged arc axis of a charged ring Given radius A charged rod has charge bent into a circular . Find electric field and has been find arc of radius and ring of charge so that point electric field , has maximum from the ring. at center of curve P. In order to solve this problem, we must Note that due to the symmetry, all of figure out an expression that expresses the terms of values cancel out. We only need , then find the maximum on the resulting graph. The x components in to sum up the of values. will cancel out, due to the symmetry of the setup. We only want want to integrate around . We don't , we want to integrate around the plane of the circle. We'll name this angle As stated by the equation must integrate over , we , or charge. It is As we go around the circle, from more favorable to integrate over angle , so that we will use linear charge density to put in terms of . to . will go . It is important to note is a constant in this situation. As stated by the equation Electric Field in must integrate over , we , or charge. It is more favorable to integrate over angle , so we will use linear charge density put in terms of (note: we must use radians for this to be in terms of to . Electric Field true) Only in the x direction As we stated before: Mushing all of these three results together yields: Only in the y direction Length As we stated before: Pythagorean theorem: Combine everything! In order to take advantage of the symmetry, we must set the horizontal as end as and the top and then double the resulting force. As complicated as this integral may appear Because of the nature by which we derived to be at first glance, one must note that the above equation, we must start using every variable presented here is a constant, radians. of course except for and Also, since , , are all constants, we can move , so we can take them out of the integral, which leaves us with: the in order to simplify the integral. It's a bit of a waste of calculus, but hey. Ah...much simpler! Now we simplify, plug, and chug. Remember equals ? Well, so we can plug in now for . Thus: , up From this, there are a few things that we can say just by looking at the nature of the equation. When field. , there is no electric also yields no electric force. When Now we substitute in (previously derived in table above) , (i.e. - the hoop appears like a point mass), , which indicates that it will act like a point charge! Here's a graph of what the function would roughly look like: How do we find that little maximum point? More calculus of course! We must take the derivative of with respect to , set it to zero, and then solve for . , right Rearrange: Product & chain rules: Rearrange: Common denominator: The part doesn't really matter so we can eliminate it: Simplify Solve for : ● Electric Flux ( ) - number of field lines intersecting a unit area. o o (dot product) This product is a minimum at and a maximum at is like the density of field lines intercepting an area, so when you multiply it by the amount of area you have, you get the # of field lines in that area. o The dot product accounts for the decrease in the amount of flux attained when the area is angled with respect to the electric field. Vector of o ● points perpendicular to the surface. It's an abstract number. Gauss' Law is used to calculate for an extended charge given certain symmetries. o Flux through a closed surface (box) ▪ If a closed surface contains no charge, the net flux ( ) through it is zero. The flux entering the surface equals the flux exiting the surface so the net flux is zero. ▪ If surface contains a dipole, the net flux ( ) is also zero since all of the field lines eventually loop around and connect again. ▪ A net enclosed charge will create a certain net flux . When charge inside is positive, the flux is positive ( ) and when the charge inside is negative, the flux is negative ( ). o To calculate net flux: o Note that for a cube, this is very hard to integrate. Thus, for our purposes, we only actually use this when the symmetries of the problem let us get constant all over the surface so , where is nice and symmetric (thus, this means using the surface area of sphere or a cylinder). Example Consider the point charge q wrapped in the Gaussian surface (the surface 1: Taking point you integrate across). charge vectors are always parallel. , the dot product goes away since o This is valid for any enclosed charge in any surface. Using this, we get the definition of Gauss' law (shown below). Thus, if you know , and the charge is arranged symmetrically, you can calculate . GAUSS' LAW: However, there are 7 special cases: (note: signifies charge enclosed by Gaussian surface) Case 1: Infinite line charge Find density segment wrapped in cylinder of length as a function of distance from wire. Consider and radius . Note: no flux is created through ends caps as the area and electric field vectors are perpendicular, thus it is ignored. & , thus , so , thus Note how is proportional to and not since we are now dealing with the electric field that originates from a line charge and not a point charge. The strength doesn't decrease over time as quickly. Poke a cylinder through it! (Gaussian cylinder). We only obtain the fluxes through the end caps. The vectors of the *Case 2: infinitely thin sides are perpendicular to the vectors of the electric field, infinitely large charged so we can ignore them. sheet, charge density It is similar to case 2, but we only get flux through top cap *Case 3: above infinite area . conducting surface, charge density We must find at . To solve, we must wrap the **Case 4: Uniform spherical object in a Gaussian sphere. charge: outside. Sphere radius and charge evenly distributed across the surface. This should look familiar. Case 5: If the sphere is conducting, The Gaussian sphere is placed inside the charged inside is zero. conducting sphere. The excess charge inside the object is zero! As it is a conducting surface, all of the extra charges repel each other and move to the surface of the sphere. , since **Case 6: inside non- conducting sphere radius with evenly distributed charge , We must use density! (charge density that is). Note that (volume charge density) is the same for Thus... and . And plugging our findings back into the original equation... And since Case 7: inside non- conducting sphere radius (given charge density - is not constant) , non uniform charge density. Solve for in sphere radius . functions of , so we must use calculus) ( and are What this means: We take an infinite number of concentric spherical shells (all parts of one particular shell have the same find since they all have the same r value), for all of them, and then add all of the s together. Also, we want to change to since it's a more convenient variable for integration. ( This is now the surface area formula since we are taking infinitely thin shell slices, which effectively have a volume equivalent to a surface area) Now... Combining with original equation ... *notes on cases 2 and 3 Look at the difference: sheet - ; conducting surface - . If we zoom out of the sheet by a lot, it begins to look like the conducting surface (remember that it does have a third dimension) BUT the sheet). In reality, is different! It is double the amount (both sides of is the same for both, but the way is obtained for both is slightly different, thus resulting in different equations. for sheet - total charge on whole disk for conducting surface - based on charge on only one surface of a larger object **implications from the resulting equations from case 4 and case 6, and . (note: indicates "proportional to") And now a summary... Gauss' Law 7 Cases General Problem Solving Steps: 1. Find through charge density ( , , or )* *way to find 1. Know for case 7 only: and thus you must use instead 2. Put in terms of 3. Substitute given equation for into equation from steps 1 & 2 4. Integrate 2. Find area of part(s) of Gaussian surface where flux is present. 3. Plug into Gauss' Law: can drop the it ) and and solve for parts and change the differential (note: for all 7 cases, you to thus just making Enclosed Charges - if you put a charge in a conducting box, you cause a separation of charge in the container. For following examples, assume the shell is neutral. at center off center inner surface charge: Charge density on inner surface is greater outer surface charge: closer to enclosed . Inside the conductor, , everywhere Electric field lines are always perpendicular to the conducting surface at equilibrium else, (the conducting shell (otherwise, the charges would move). effectively causes a gap in the field of the Inside conducting shell, conductor) On outer surface, charge is arranged Negative charge is distributed uniformly on uniformly. There is no information about the inner surface and positive charge is distributed uniformly on the outer surface. position of inner . In order to obtain the outside charge, just add the charge of the enclosed charge and the charge of the shell. Note that if you have a charge of value charge ● , and you have a shell of outside the shell. Conductor in an intensity, the field. - Similar to how sun rays hit the equator with higher field lines will hit the equator of the conducting sphere with a higher flux, causing more charge to accumulate there. ● Conducting Sheets o 1 thin conducting sheet of area , charge , , so thus o 2 parallel plates - It is essentially the same as adding the electric field of both plates in isolation. The field on the right is really a combination of the rigtwardly headed field lines of the left plate and the right plate, and the field on the left is really a combination of the leftwardly headed field lines of the left plate and the right plate. The field lines in the middle are heading opposite directions with the same magnitude and thus cancel out. Thus the field on the outside is , which is twice as much as the field on one side of the plate if it was isolated.