Guest Zugravu Moni Posted March 21, 2005 Posted March 21, 2005 Could you tell me how to compute the electric field strength at the edge of a disk and the electric field strength of a ultra-long wire? Thanks!
swansont Posted March 21, 2005 Posted March 21, 2005 Could you tell me how to compute the electric field strength at the edge of a disk and the electric field strength of a ultra-long wire? Thanks! For the wire you have a symmetry that will allow you to use Gauss's law.
Johnny5 Posted March 21, 2005 Posted March 21, 2005 Could you tell me how to compute the electric field strength at the edge of a disk and the electric field strength of a ultra-long wire? Thanks! I will do the ultra long wire, it's been awhile since I've done it, so i am a bit out of practice. Start off with the definition of an electric field: [math] \vec E = \frac{1}{4\pi\epsilon_0} \int \lambda dl \frac{\hat R}{R^2} [/math] In the formula above, lambda l is the linear charge density, which will have units of coulombs per meter. Suppose that you have a line of finite length L, with electrons situated along the length, in regular intervals, so that the electric charge per unit length is constant. Assume the distance between the electrons is very small, so that we can treat lambda as continuous (this assumption justifies the use of the integral). Now, choose a reference frame to analyse the wire in. Let your frame be as follows: Let the origin of your reference frame be at the center of the wire. Let the wire lie on the X axis of your frame. Thus, L/2 of the wire lies on the positive X axis, and L/2 of the wire lies on the negative X axis. R vector is defined by the following vector triangle. [math] \vec r^\prime + \vec R = \vec r [/math] Where [math] \vec r [/math] Is the vector from the origin of your reference frame, to an arbitrary field point (x,y,z). And with [math] \vec r^\prime [/math] denoting a vector from the origin of your reference frame, to an electron in the wire, at location (x`,y`,z`). Thus, you can think of the unprimed position vector r as locating a point in the vacuum, and the primed position vector as going into some electrically charged material body. Each of these two vectors r,r` are position vectors, which means that they start from point (0,0,0) and go elsewhere. Suppose that this line charge is the only thing in the universe, so that if it didn't exist, the magnitude of the electric field E, at the field point (x,y,z) would be given by: [math] | \vec E(x,y,z) | = 0 [/math] Let's just do the part of the integral from x=0, to x=L/2 thus, dl=dx. Therefore, we have the following integral: [math] \vec E = \frac{1}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} \lambda dx \frac{\hat R}{R^2} [/math] Since lambda is constant, we can pull it out of the integral sign, to obtain the following: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx \frac{\hat R}{R^2} [/math] Using the equation which defines R vector we have: [math] \vec R = \vec r - \vec r^\prime [/math] In rectangular coordinates we have: [math] \vec r = x \hat i + y \hat j + z \hat k [/math] [math] \vec r^\prime = x^\prime \hat i + y^\prime \hat j + z^\prime \hat k [/math] Since the line charge is positioned entirely along the x axis of the frame, it follows that for any electron in the wire, y`=0, and z`=0, therefore: [math] \vec r^\prime = x^\prime \hat i [/math] Thus, we have: [math] \vec R = (x-x^\prime) \hat i + y \hat j + z \hat k [/math] The magnitude of this vector can be found using the Pythagorean statement. Hence: [math] R = |\vec R| = \sqrt{ (x-x^\prime)^2 + y^2 + z^2 }[/math] Now, vector R is equal to its magnitude times its direction, symbolically: [math] \vec R = R \hat R [/math] From which it follows that: [math] \hat R = \frac{\vec R}{R} [/math] The previous relation allows us to write our work thus far, as follows: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx \frac{\vec R}{R^3} [/math] In order to differentiate with respect to x, we need to replace R vector and magnitude of R vector, by equivalent functions of x coordinate. Therefore: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx \frac{[(x-x^\prime) \hat i + y \hat j + z \hat k]}{R^3} [/math] [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx \frac{[(x-x^\prime) \hat i + y \hat j + z \hat k]}{[(x-x^\prime)^2 + y^2 + z^2]^{3/2}}[/math] Now, we can use the fact that the integral of a sum, is the sum of the integrals, and perform three separate integrals, one of which will give us the component of E in the i^ direction, another of which will give us the component of E in the J^ direction, and another of which will give us the component of E in the k^ direction. Or course we have neglected contributions to the total electric field at (x,y,z) due to the electrons on the negative x axis, but we can take them into account later. The first integral is as follows: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx \frac{(x-x^\prime) \hat i }{[(x-x^\prime)^2 + y^2 + z^2]^{3/2}} [/math] Let [math] U^2 = (x-x^\prime)^2 + y^2 + z^2 [/math] This simplifies the appearance of the integral a bit. We now have: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx \frac{(x-x^\prime) \hat i }{U^3} [/math] Now, differentiate both sides of the definition of U^2 treating all symbols other than x as constants, to obtain: [math] 2UdU = 2(x-x^\prime) dx [/math] From which it follows that: [math] UdU = (x-x^\prime) dx [/math] From which it follows that: [math] UdU \hat i = (x-x^\prime) dx \hat i[/math] We can now write the integral as follows: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dU \frac{U \hat i }{U^3} [/math] Now, the reference frame isn't spinning, so that the direction of i^ is constant in time (most importantly it's not a function of x), so we can pull the direction out of the integral, and we can cancel the U in the numerator, with a U in the denominator to obtain: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i \int_{x=0}^{x=L/2} dU \frac{1}{U^2} [/math] And since [math] \frac{1}{U^2} = U^{-2} [/math] We can write the integral as follows: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i \int_{x=0}^{x=L/2} U^{-2} dU [/math] Theorem: [math] \int_a^b x^n dx = [\frac{x^{n+1}}{n+1}]_a^b [/math] Using the previous theorem, we can solve the integral. [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [-\frac{1}{U}]_{x=0}^{x=L/2} [/math] We can eliminate the minus sign if we flip the endpoints of integration, therefore we have: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [\frac{1}{U}]^{x=0}_{x=L/2} [/math] Now, we have two choices. We can either replace U by the definition of it in terms of x, or we can rewrite the limits of integration in terms of U. [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [\frac{1}{\sqrt{ (x-x^\prime)^2 + y^2 + z^2} }]^{x=0}_{x=L/2} [/math] So we have: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [\frac{1}{\sqrt{x^\prime^2 + y^2 + z^2}} - \frac{1}{ \sqrt{(L/2-x^\prime)^2 + y^2 + z^2}} }] [/math] Now, we have two more integrals to do, but from a computational standpoint, they are identical. We now have to integrate the following: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} \frac{y \hat j dx}{[(x-x^\prime)^2 + y^2 + z^2]^{3/2}}[/math] The integral is with respect to x, so that y j^ is constant during the integral, therefore we can pull it out of the integral, to obtain: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac{dx}{[(x-x^\prime)^2 + y^2 + z^2]^{3/2}}[/math] Let us make the same substitution as before, so let [math] U^2 = (x-x^\prime)^2 + y^2 + z^2 [/math] So we are trying to integrate this: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac{dx}{U^3}[/math] We also have: [math] UdU = (x-x^\prime) dx [/math] From which it follows that: [math] dx = \frac{UdU}{ (x-x^\prime)} [/math] And using the definition of U, we have: [math] (x-x^\prime) = \sqrt{U^2 -(y^2+z^2)} [/math] Therefore: [math] dx = \frac{UdU}{ \sqrt{U^2 -(y^2+z^2)}} [/math] Thus, the integral we are trying to evaluate is: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac{dU}{ U^2 \sqrt{U^2 -(y^2+z^2)}} [/math] This isn't an integral which I know right away, so you could use a table of integrals. I am going to try to make a trig substitution though. y and z are constant during the integration, so lets introduce the constant a as follows: [math] a^2 = y^2+z^2 [/math] So we have: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac{dU}{ U^2 \sqrt{U^2 - a^2}} [/math] Now, imagine a right triangle, with hypotenuse U, one side of length a, and the other side of length b. Therefore, we have: [math] a^2+b^2 = U^2 [/math] From which it follows that: [math] b = \sqrt {U^2 - a^2} [/math] So we can rewrite the integral as: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac{dU}{ U^2 b} [/math] Introduce angle q so that: [math] b=U cos \theta [/math] Therefore: [math] U = b sec \theta[/math] Changing the integral a bit more gives us: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac{dU}{ b^3 sec^2 \theta} [/math] To finish changing from an integral over U, to an integral over theta, we have to figure out what the differential of secant q is, because [math] dU = b d [sec \theta ] [/math] In fact, let us make the substitution now, so that we have: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac{d [sec \theta ]}{ b^2 sec^2 \theta} [/math] So what is the differential of secant theta? Or equivalently, what is the differential of 1/cosine theta? We can use the quotient rule of the differential calculus. [math] d[sec\theta] = d[ \frac{1}{cos \theta} ] = \frac{0+sin \theta}{cos^2 \theta} d\theta = (\frac{1}{cos\theta})(\frac{sin\theta}{cos\theta}) d\theta = sec \theta tan \theta d \theta [/math] Therefore, we have to perform the following integral: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac {sec \theta tan \theta d \theta}{ b^2 sec^2 \theta} [/math] Which simplifies slightly to: [math] \vec E = \frac{\lambda}{4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac {tan \theta d \theta}{ b^2 sec \theta} [/math] b is a constant, so we might as well pull that out of the integral, so we have: [math] \vec E = \frac{\lambda}{b^2 4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} \frac {tan \theta d \theta}{ sec \theta} [/math] Since cosine is one over the secant, and tangent is sine divided by cosine, the integral simplifies to: [math] \vec E = \frac{\lambda}{b^2 4\pi\epsilon_0} y \hat j \int_{x=0}^{x=L/2} sin \theta d \theta [/math] The integral of sine theta is -cosine theta, so that we have: [math] \vec E = -\frac{\lambda}{b^2 4\pi\epsilon_0} y \hat j cos \theta |_{x=0}^{x=L/2} [/math] Now, we have to rewrite cosine theta in terms of x. [math] cos \theta = \frac{b}{U} [/math] U in terms of x is: [math] U = \sqrt{(x-x^\prime)^2 + y^2 +z^2} [/math] So therefore cosine theta in terms of x is given by: [math] cos \theta = \frac{b}{\sqrt{(x-x^\prime)^2 + y^2 +z^2}} [/math] So the j^ component of the electric field at position (x,y,z) in the reference frame, due to electrons from x=0, to x=L/2 in the wire is given by: [math] \vec E = -\frac{\lambda}{b^2 4\pi\epsilon_0} y \hat j \frac{b}{\sqrt{(x-x^\prime)^2 + y^2 +z^2}} |_{x=0}^{x=L/2} [/math] We can get rid of the negative sign, if we flip the endpoints of integration, therefore: [math] \vec E = \frac{\lambda}{b^2 4\pi\epsilon_0} y \hat j \frac{b}{\sqrt{(x-x^\prime)^2 + y^2 +z^2}} |^{x=0}_{x=L/2} [/math] And evaluating at the endpoints gives: [math] \vec E_2 = \frac{\lambda}{b 4\pi\epsilon_0} y \hat j [ \frac{1}{\sqrt{x^\prime^2 + y^2 +z^2}} - \frac{1}{\sqrt{(L/2-x^\prime)^2 + y^2 +z^2}} ] [/math] And we can write the solution for E3 immediately, since the integral would have been identical, so we have: [math] \vec E_3 = \frac{\lambda}{b 4\pi\epsilon_0} z \hat k [ \frac{1}{\sqrt{x^\prime^2 + y^2 +z^2}} - \frac{1}{\sqrt{(L/2-x^\prime)^2 + y^2 +z^2}} ] [/math] So the electric field at the field point (x,y,z) due to all the electrons on the postive x axis is given by: [math] \vec E = \vec E_1 +\vec E_2 + \vec E_3 [/math] Now, rather than go back and do more work, let us have done the analysis in a reference frame in which one end of the wire was located at the origin, and the other end was located at a distance L, on the positive X axis. Thus, the solution we already obtained holds by simply replacing L/2 by L, in E1,E2,E3 above. However, something to be more concerned about, is whether or not we introduced a minus sign error, by letting x vary in the integral, rather than the source coordinate x`. I will now solve the problem using Gauss' law, take the appropriate limits, and compare the answers.
swansont Posted March 21, 2005 Posted March 21, 2005 I think that's pretty good motivation to use Gauss's law.
Johnny5 Posted March 21, 2005 Posted March 21, 2005 I think that's pretty good motivation to use Gauss's law. In the end I will use Gauss' law too, unless you beat me to it.
swansont Posted March 21, 2005 Posted March 21, 2005 In the end I will use Gauss' law too, unless you beat me to it. The goal isn't to do the problem for the OP, but to help him or her do the problem. Especially when it quacks like a homework problem.
Johnny5 Posted March 21, 2005 Posted March 21, 2005 The goal isn't to do the problem for the OP, but to help him or her do the problem. Especially when it quacks like a homework problem. Oh, I see. I didn't see myself as solving someone's homework for them, but there are any number of places which already have the answer right? Isn't it highly instructive to actually demonstrate as detailed a solution as possible? Besides I could make an error, such as letting x vary, instead of x`, for example. I was waiting for someone to say something. The general approach to solution of EM problems in particular is a skill, and I think that's what I wanted to show them, more than solve their homework problem. I will keep that in mind next time. Perhaps solve a similiar problem, so that I don't end up doing someone's homework for them.
Johnny5 Posted March 21, 2005 Posted March 21, 2005 Suppose you are asked to find the electric field of an infinitely long wire, instead of an Ultra-long wire, at a field point (x,y,z). You can use Gauss' law for this. Gauss' Law [math] \Phi_e = \oint \vec E \bullet d\vec a = \frac{Q_{enc}}{\epsilon_0} [/math] Let the infinitely long wire coincide with the x axis of the reference frame. Now, before using Gauss' law think about the electric field at an arbitrary field point (x,y,z). [math] \vec E(x,y,z) = \vec E_1 \hat i + \vec E_2 \hat j + \vec E_3 \hat k [/math] Suppose that we already had the answer for the electric field at point (x,y,z) in the frame, and that there is an electron at that point. The electron will be subjected to a force, therefore it will begin accelerating in the frame. The force is equivalent to the product of the electric field, and the electric charge. That is: [math] \vec F = q \vec E [/math] Where q is the electric charge of the test particle, in this case an electron. Let -e denote the electric charge of an electron. Therefore, that electron, which is located at field point (x,y,z) will be subjected to the following force: [math] \vec F = -e \vec E [/math] Since the electric field at (x,y,z) is due to electrons (rather than protons) located on the x axis, we should expect the force to be repulsive, rather than attractive. Therefore, we should expect the electron to be pushed away from the wire. It can now be argued that there will be no force component in either the i^ direction, or -i^ direction. Suppose the electron is initially located at (0,1,0). The electrons on the positive axis will push the electron diagonally, into the second quadrant of the XY plane, BUT, there are just as many electrons on the negative X axis, and they will push the electron diagonally, into the first quadrant of the XY plane just as hard as the electrons on the positive X axis push it, and so the force components in the i^ direction and -i^ direction will cancel. But there will be equal contributions to the net force in the j^ direction, from the electrons on the positive x axis, and negative x axis, so we can anticipate that the net force on the electron at (0,1,0) will be purely in the j^ direction. Now, if we choose our Gaussian surface to be a cylinder, with the line charge running along the axis of the cylinder, the electric field at a surface point on the cylinder will point in a direction parallel to the outward normal of a differential area. Thus, we have: [math] \oint \vec E \bullet d\vec a = \oint \vec E da [/math] Because of symmetry, E is constant along the surface of they cylinder. There is no reason that E would be stronger at one point on the surface of the cylinder than any other. Therefore we can pull E outside of the integral to obtain: [math] E \oint da = \frac{Q_{enc}}{\epsilon_0} [/math] Let the Gaussian cylinder have radius R, and length L. Therefore the surface area of it is [math] 2\pi R L [/math] So that we have: [math] E 2\pi R L = \frac{Q_{enc}}{\epsilon_0} [/math] Now, we just have to figure out the amount of charge enclosed by the Gaussian surface Qenc. We know the charge per unit length is l and that it is constant over any wire length L, so that the total charge enclosed is [math] \lambda L [/math] Therefore: [math] E 2\pi R L = \frac{\lambda L}{\epsilon_0} [/math] We can now solve for the electric field strength at a point on the surface of our Gaussian cylinder. We have: [math] E = \frac{\lambda }{2\pi R \epsilon_0} [/math] This result says that the electric field of the wire is inversely proportional to the perpendicular distance you are away from the wire. Now, we can take the limit as the length of the finite wire goes to infinity, in the longer problem above. If we don't get the answer which Gauss' law gave us, then there was a mistake somewhere, which hopefully will be easily locatable.
Johnny5 Posted March 22, 2005 Posted March 22, 2005 In our first attempt to solve for the electric field of a wire of finite length, we obtained a very complicated formula, but had no way to check to see if the formula made sense. Here was the formula obtained: [math] \vec E = E_1 \hat i + E_2 \hat j + E_3 \hat k [/math] Where [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [\frac{1}{\sqrt{x^\prime^2 + y^2 + z^2}} - \frac{1}{ \sqrt{(L/2-x^\prime)^2 + y^2 + z^2}} }] [/math] [math] \vec E_2 = \frac{\lambda}{b 4\pi\epsilon_0} y \hat j [ \frac{1}{\sqrt{x^\prime^2 + y^2 +z^2}} - \frac{1}{\sqrt{(L/2-x^\prime)^2 + y^2 +z^2}} ] [/math] [math] \vec E_3 = \frac{\lambda}{b 4\pi\epsilon_0} z \hat k [ \frac{1}{\sqrt{x^\prime^2 + y^2 +z^2}} - \frac{1}{\sqrt{(L/2-x^\prime)^2 + y^2 +z^2}} ] [/math] Let us at least send the end of the wire on the positive x axis, to positive infinity, so that we are taking the limit as L goes to infinity of the forumla above. After taking the limit we have: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [\frac{1}{\sqrt{x^\prime^2 + y^2 + z^2}}] [/math] [math] \vec E_2 = \frac{\lambda}{b 4\pi\epsilon_0} y \hat j [ \frac{1}{\sqrt{x^\prime^2 + y^2 +z^2}} ] [/math] [math] \vec E_3 = \frac{\lambda}{b 4\pi\epsilon_0} z \hat k [ \frac{1}{\sqrt{x^\prime^2 + y^2 +z^2}} ] [/math] Now, we already used Gauss' law to figure out that the electric field of an infinite wire has no component in the i^ direction, but the formula above does have a component in the i^ direction, however, we did not take the limit as as both ends of the wire approach infinity yet. But look what happens if we combine E2,E3 as follows: [math] E_2 \hat j + E_3 \hat k = \frac{\lambda}{b 4\pi\epsilon_0} [ \frac{y \hat j + z \hat k}{\sqrt{x^\prime^2 + y^2 +z^2}} ] [/math] Now, suppose that we made a mistake by integrating over the field point x, instead of the source point x`. Then the forumla above would be: [math] E_2 \hat j + E_3 \hat k = \frac{\lambda}{b 4\pi\epsilon_0} [ \frac{y \hat j + z \hat k}{\sqrt{x^2 + y^2 +z^2}} ] [/math] Which could then be written as: [math] E_2 \hat j + E_3 \hat k = \frac{\lambda}{b 4\pi\epsilon_0} [ \frac{y \hat j + z \hat k}{R} ] [/math] Now, recall that: [math] b = \sqrt{U^2 - y^2 - z^2 } [/math] And [math] U^2 = (x-x^\prime)^2 + y^2 + z^2 [/math] From which it follows that: [math] b = \sqrt{(x-x^\prime)^2} [/math] Therefore: [math] E_2 \hat j + E_3 \hat k = \frac{\lambda}{\sqrt{(x-x^\prime)^2} 4\pi\epsilon_0} [ \frac{y \hat j + z \hat k}{R} ] [/math] Now, look at E1. [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [\frac{1}{\sqrt{x^\prime^2 + y^2 + z^2}}] [/math] Suppose that we place an electron at the origin (0,0,0). There are an infinite number of electrons along the positive x axis, all pushing the electron in the -i^ direction, and nothing is pushing it in any other direction. In the formula above, we have division by zero error. We can eliminate the division by zero error, if we place the electron someplace on the negative X axis. So let us place a test charge (an electron) at (-1,0,0). Intuitively, all the electrons which are on the positive x axis now push the test charge in the -i^ direction, lets see what the formula predicts: The first problem we encounter, is that we cannot put in the location of the test particle, unless we assume that x`=x. Even if that is the case we get this: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \hat i [/math] So the electric force on the test charge would then be given by: [math] \vec F = -e\vec E_1 = -\frac{e\lambda}{4\pi\epsilon_0} \hat i [/math] But from the formula for E2+E3 above, the case where x=x` leads to division by zero error (if we release the test charge at (-1,1,0) for example), and the total force on the test charge is not just E1, it is E1+E2+E3. The point is, we wanted to find E(x,y,z), not E(x`,y,z), and the only way to fix the problem is to go back and perform the integral over changes in the source coordinates. When I first studied EM, I couldn't understand why it mattered. It seemed to me, that having E(x,y,z) allowed to you to evaluate the Electric field due to some charge distribution at any point in space, interior or exterior to a charged body, so i didn't understand why I couldn't perform the integral allowing the field point to vary. However, the answers that you get if you perform the integral with the field point varying instead of the source point, give physically unrealistic solutions. Way way back, when we first started answering the problem, we got: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx \frac{(x-x^\prime) \hat i }{[(x-x^\prime)^2 + y^2 + z^2]^{3/2}} [/math] Right there is where an error was made. We want the integral to be taken over the source points, not the field points. In other words, we want the source coordinates to vary in the integral. So what should have been written was this: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} dx^\prime \frac{(x-x^\prime) \hat i }{[(x-x^\prime)^2 + y^2 + z^2]^{3/2}} [/math] We will get almost the same answer as before, the only difference will be a minus sign. So we try the integral again. Let U2 be defined as follows: [math] U^2 = (x-x^\prime)^2 + y^2 + z^2 [/math] Now, during the integral, we think of the point (x,y,z) as fixed, and the field point (x`,y`,z`) as varying. So differentiating we have: [math] 2UdU = -2(x-x^\prime) dx^\prime [/math] Which differs from what we got before by a minus sign. Therefore: [math] -UdU \hat i = (x-x^\prime) dx^\prime \hat i [/math] Making the appropriate substitutions we have: [math] \vec E_1 = \frac{\lambda}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} \frac{-UdU \hat i}{U^3} [/math] So [math] \vec E_1 = -\frac{\lambda \hat i}{4\pi\epsilon_0} \int_{x=0}^{x=L/2} \frac{dU}{U^2} [/math] After integrating we have: [math] \vec E_1 = -\frac{\lambda \hat i}{4\pi\epsilon_0} [- \frac{1}{U} ] _{x=0}^{x=L/2} [/math] So we have: [math] \vec E_1 = \frac{\lambda \hat i}{4\pi\epsilon_0} [\frac{1}{\sqrt{(x-x^\prime)^2 + y^2 + z^2 }} ] _{x=0}^{x=L/2} [/math] Now, the endpoints of integration go from x`=0 to x`=L/2, since the source coordinate was varying. Therefore, we have: [math] \vec E_1 = \frac{\lambda \hat i}{4\pi\epsilon_0} [\frac{1}{\sqrt{(x-L/2)^2 + y^2 + z^2 }} - \frac{1}{\sqrt{x^2 + y^2 + z^2 }} ] [/math] So this gives the contribution to a nonzero electric field at a field point (x,y,z), due to all the electrons on the positive x axis, in the i^ direction. This time, lets take into account contributions from the electrons on the negative x axis as well. The integral would have thus gone from x`=-L/2 to x`=L/2, so that we get this: [math] \vec E_1 = \frac{\lambda \hat i}{4\pi\epsilon_0} [\frac{1}{\sqrt{(x-L/2)^2 + y^2 + z^2 }} - \frac{1}{\sqrt{(x+L/2)^2 + y^2 + z^2 }} ] [/math] Now, consider test charges located entirely in the YZ plane, therefore their X coordinate must be zero. For any electron located in the YZ plane, we have: [math] \vec E_1 = \frac{\lambda \hat i}{4\pi\epsilon_0} [\frac{1}{\sqrt{(L/2)^2 + y^2 + z^2 }} - \frac{1}{\sqrt{(L/2)^2 + y^2 + z^2 }} ] = 0 [/math] We finally get the intuitive result that an electron located directly above the midpoint of a finite wire of length L will not be pushed to the left or right, its going to be pushed directly away from the wire, as we shall soon see. And in the case of the infinitely long wire, we get E1=0 no matter what the x coordinate of the test charge is. The only thing remaining now, is to compute E2,E3 correctly, then put it all together, and we will have solved for the electric field of a wire of finite length L, and constant charge density l in a frame in which the center of the wire is located at the origin, and the wire itself lies on the x axis of the frame, and additionally the wire (and electrons) is at rest in the frame.
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