Step 1: Define the linear mass density of the rod. Then we can express the left-hand side in explicit for as E dA cosine of 0. A rod of length L lies along the x-axis with its left end at the origin. Question . - YouTube A rod of length L lies along the x-axis with its left end at the origin. By using the same procedure, we calculate the amount of charge along the next shell, and then the next shell, and then we add all those charges to one another. As little r changes, then the will also change. (2DEG) in the drift region between the gate and the drain that has a non-uniform lateral 2DEG distribution that increases in a direction in the drift region from the gate to the drain. Step 2: Replace dm in the definition . #1 The plastic rod of length \displaystyle L L in the diagram has non uniform linear charge density \displaystyle \lambda = cx =cx where c is a positive constant. This is the amount of charge distributed through this incremental spherical shell. Here again, 4 and R and s, these are all constant, so we can take it outside of the integral. First, lets try to figure out the total charge of the distribution as the question mark. Here we will integrate this from 0 to little r. Thats our region of interest to be able to get the q-enclosed, the net charge inside of the volume surrounded by the Gaussian sphere. The integral on the right-hand side goes along the length of the rod, from x= to x=9.8 m. 2022 Physics Forums, All Rights Reserved, Electric field of infinite plane with non-zero thickness and non-uniform charge distribution, E-field of solid sphere with non-uniform charge density, Electric Field of a Uniform Ring of Charge, Calculation of Electrostatic Potential Given a Volume Charge Density, The potential electric and vector potential of a moving charge, Uniform charge density and electric potential, Find the Electric potential from surfaces with uniform charge density, Electrostatic potential energy of a non-uniformly charged sphere, Charge density on the surface of a conductor, Potential on the axis of a uniformly charged ring, Problem with two pulleys and three masses, Newton's Laws of motion -- Bicyclist pedaling up a slope, A cylinder with cross-section area A floats with its long axis vertical, Hydrostatic pressure at a point inside a water tank that is accelerating, Forces on a rope when catching a free falling weight. Lets not forget the r2 term that we had left from this step, so we will have an r2 over here. uniform distribution is red; non-uniform is blue uniform distribution is blue; non-uniform is red not enough information is given to say This particular non-uniform distribution has less charge in the center and more concentrated toward the outside of the sphere than the uniform distribution has. It may not display this or other websites correctly. Transcribed Image Text: Q1 A rod carrying a non-uniform linear charge density (2 = ax) lies along A the positive x-axis with its left end at a distance (a) from the origin. dV is in volts, not in meters. Point A lies on the x-axis a distance 3.59 m to the left of the rod, as shown in the figure. Therefore we will have Q over 0 on the right-hand side, and solving for electric field we will have Q over 40r2. On the right-hand side, we will have q-enclosed over 0. You are using an out of date browser. A non-conducting rod of length l with a uniform charge density and a total charge Q is lying along the x -axis, as illustrated in the figure. the two relevant equations i can find are: wait so then what would be the values for the r? Join / Login >> Class 11 >> Physics >> Systems of Particles and Rotational Motion >> Centre of Mass of Different Objects . If it is not necessary dont use it. Charge is distributed non-uniformly throughout the volume of the distribution, which has radius of big R, and the charge density was given as a constant s times little r over big R, and little r is the location of the point of interest. The function ( r, , y) is the density function. For dq, which is associated with the volume of this incremental, spherical shell, and the volume of that is surface area, 4s2, times its thickness, ds. Example 4: Electric field of a charged infinitely long rod. The integral of dV is simply V(A) the potential at A. The rod has a non-uniform linear charge density = x, where = 0.009 C/m2 and x is the position. Show that the electric field E at point P a distance R above one end of the rod makes an angle of 45 with the rod and that this result is independent of the distance R. Homework Equations $$\vec{E}=\int \frac{k\lambda }{r^2}dx$$ The Attempt at a Solution (2DEG) in the drift region between the gate and the drain that has a non-uniform lateral 2DEG distribution that increases in a direction in the drift region from the gate to the drain. . Q is the total charge on the rod. Now, were going to consider an example such that the charge density is not constant. Click hereto get an answer to your question The density of a non - uniform rod of length 1m is given by (x) = a(1 + bx^2) where a and b are constants and 0 < x < 1 . Draw a picture.2. Then we will add all of those dqs to one another throughout the region inside of this Gaussian sphere. The rod of non-uniform linear charge density A = axe (where a = 3.00 nC/m3) placed on the x-axis such a way that one of its ends is at the origin and the other end is at 0.4 m. Find the electric potential on the y-axis at (0 m, 0.3 m). In other words, even before we apply these steps, we can say that the system will behave like a point charge and total electric field is going to be equal to this quantity. Therefore in explicit form this is going to be equal to charge density, s, times s over R. This is charge per unit volume, times the volume of the region that were interested with. Consider a rod in three dimensional space where y is the height axis. So assume there is an insulated sphere with a non-uniform charge density and radius R. It has a constant electric field of E. Here is my current line of thinking: We can pick a Gaussian surface at radius r < R. That would give E ( 4 r 2) = q ( r) o, where q ( r) is a function which defines the charge enclosed by the Gaussian surface. As shown in the figure, a rod of length 9.8 m lies along the x-axis, with its left end at the origin. Both of them are correct answers for this case. Remember that we have found in the previous part, that the Q was equal to sR3. In order to get the q-enclosed throughout this region, which is the region surrounded by Gaussian sphere, as in the previous part of this problem, we are going to choose an incremental spherical shell at an arbitrary location, lets say something like this. 1.2K subscribers This is an example of using calculus to find the electric potential of a continuous charge distribution, in this case for a rod with a non-uniform linear charge. The Coulomb constant is 8.988 109 N m2/C2. Physics | Electrostatics | Non Uniformly Charged Rod | by Ashish Arora (GA) 4,399 views Nov 26, 2015 http://www.physicsgalaxy.com Learn complete Physics Video Lectures on Electrostatics for. If we had uniform charge distribution throughout the volume of this distribution, as we had done in one of the earlier examples, we would have expressed the volume charge density of the distribution, and simply by taking the product of that density with the volume of that region that were interested in, which is the region inside of the Gaussian sphere, we would have ended up with the q-enclosed, the net charge inside of this region. We have a solid, spherical charge distribution charge is not distributed uniformly throughout the volume of this object such that its volume charge density varies with is equal to s times r over R. So in other words, as we go away from the center of the distribution, which has a radius of big R, as we move radially out from the center of the distribution, the charge density increases. The centre of mass of the rod will be at: As a first example for the application of Coulomb's law to the charge distributions, let's consider a finite length uniformly charged rod. ok, so then in the last integral you will have a ln, but would that be the same that you integrate over? Write dq in terms of the geometry (and the charge density).4. (a) What is the magnitude of the electric field from the axis of the shell? Well, of course if the charge were distributed uniformly and therefore the charge per unit volume would have been the same at every point inside of this region, and to be able to get the total charge of the distribution, we would have directly taken the product of the volume charge density by the volume of the whole distribution, which would have given us the total charge. So, the left-hand side of the Gausss law is identical to the previous spherical symmetry problems. Then the final expression for the electric field is going to be, in terms of the total charge of the distribution inside of the sphere, as Q over 40R4 times r2. In this notation then, in the numerator, we will have the total charge of the distribution. h is the height of the rod and a is the radius of the rod. A NON-UNIFORM LATERAL PROFILE OF TWO-DIMENSIONAL ELECTRON GAS CHARGE DENSITY IN TYPE III NITRIDE HEMT DEVICES USING ION IMPLANTATION THROUGH GRAY SCALE MASK. Hint: This exercise requires an integration. When we look at that region, we see that it encloses all the charge distributed throughout the sphere. This problem has been solved! Part A What is the total charge on the rod? That is the electric field generated by this charge distribution at a point outside, r distance away from the center of the distribution. The rod has a non-uniform linear charge density = x, where = 0.009 C/m2 and x is the position. In other words, charge density was constant throughout the distribution. The incremental amount of charge that is distributed throughout this incremental spherical shell will be equal to times the volume of incremental shell. He has created a youtube channel in the name of Physics Galaxy. To keep yourself updated about physics galaxy activities on regular basis follow the facebook page of physics galaxy at https://www.facebook.com/physicsgalaxy74The website, aimed at nurturing grasping power students, has classroom lectures on almost all the topics. The positively charged rod has total charge Q. The left-hand side will be identical to the previous part, which will eventually gives us electric field times the surface area of the sphere, which is 4r2, and the q-enclosed in this case is the net charge inside of the region surrounded by this Gaussian sphere. Now let us look at the electric field outside of this distribution for r is larger than R. The simplest way of handling such a problem is since we are dealing with a spherical charge distribution with radius big R and we are interested in the electric field at a point outside of the distribution, again applying Gausss law, we simply place a Gaussian sphere using the spherical symmetry passing through the point of interest. (a) With V = 0 at infinity, find the electric potential at point \displaystyle P_2 P 2 on the y-axis, at a distance y from one end of the rod. Example 6- Electric field of a non-uniform charge distribution. Compute the electric field at a point P, located at a distance y off the axis of the rod. Apply this to known results for dV due to a small charge dq (like Coulomb's law).5. The mass of the rod can be calculated with. Find the potentia at A, and answer in units of volts This is in radial direction, so we can multiply this by the unit vector pointing in radial direction in order to express the electric field in vector form. Example: Infinite sheet charge with a small circular hole. In doing that, we will have E is equal to sR3, multiplying them by R3, and also multiply the denominator by the same quantity in order to keep the ratio unchanged. A "semi-infinite" nonconducting rod has a uniform linear charge density . A non-conducting rod of length l with a uniform charge density and a total charge Q is lying along the x -axis, as illustrated in the figure. If that is the case, then this will allow us to be able to calculate the amount of charge associated with this incremental shell. Adding all the incremental area vectors along this surface, we will eventually end up with the surface area of that sphere, which is going to be 4r2. Question: A rod has a non-uniform charge density 1 = Bx2 where B is a constant, and endpoints at (L,0) and (2L,0). So q-enclosed is going to be equal to s over R times r to the 4th. This result is for the case that the point of interest is inside of the distribution. Now we will look at the right-hand side. In other words, its going to be equal to 4r2 times dr. I'm not sure I understand why I need to use ##d##.. If we do that, we will have E times 4r 2 is equal to s over big R times little r4 divided by 0 on the right-hand side of the Gausss law expression. The rod is coaxial with a long conducting cylindrical shell (inner radius=5.0 cm , outer radius=10 cm ). The net charge on the shell is zero. The distance of centre of mass of rod from the origin is: Solve Study Textbooks Guides. If we integrate this quantity here, s 4 and big R, these are all constant, we can take it outside of the integral. Besides uploading transcripts of all his videos, he has created a software based synchronized European voice accent of all videos to benefit students in USA, Europe and other countries.Reference link of this video is at https://youtu.be/i5IId5voOA4 In order to have the same relationship, lets multiply both the numerator and denominator of this expression by R3. It has a non uniform charge density $\\lambda = \\alpha x$ where $\\alpha$ is a positive . Eventually we add all those incremental charges to one another throughout the volume of this whole distribution and get the total charge. . http://www.physicsgalaxy.com Learn complete Physics Video Lectures on Electrostatics for IIT JEE by Ashish Arora. Introductory Physics Homework Help Potential due to a rod with a nonuniform charge density archaic Sep 15, 2020 Sep 15, 2020 #1 archaic 688 210 Homework Statement: A rod of length lies on the x-axis such that its left tip is at the origin. A non-uniform thin rod of length L is placed along x . This is the most comprehensive website on Physics covering all the topics in detail. dV is integrated over the potential belonging to zero charge to the actual potential V(A) of the total charge of the rod. Here, we can do some cancellations, dividing both sides by r2, we are going to end up with r2 on the right-hand side and we can cancel s here. In this case, we cannot do that, we cannot take the product of charge density with the volume of whole distribution to be able to get the total charge because is not the same at every point inside of this region. The example illustrates a general strategy for solving problems of this type:1. Now we can try to express this in terms of the total charge of the distribution. Also, if you recall, we said that whenever we are dealing with spherical charge distributions, then for all exterior points, the system behaves as if all the charge is concentrated to its center, and it behaves like a point charge for all of the exterior points, therefore the problem reduces to a point charge problem, such that we are calculating the electric field that it generates at point P, which is r distance away from the charge, generating an electric field in a radially outward direction exactly like in this case. from Office of Academic Technologies on Vimeo. You are using an out of date browser. This is an example of using calculus to find the electric potential of a continuous charge distribution, in this case for a rod with a non-uniform linear charge density. For a better experience, please enable JavaScript in your browser before proceeding. Example 1: Electric field of a point charge, Example 2: Electric field of a uniformly charged spherical shell, Example 3: Electric field of a uniformly charged soild sphere, Example 4: Electric field of an infinite, uniformly charged straight rod, Example 5: Electric Field of an infinite sheet of charge, Example 6: Electric field of a non-uniform charge distribution, Example 1: Electric field of a concentric solid spherical and conducting spherical shell charge distribution, Example 2: Electric field of an infinite conducting sheet charge. A rod of length L has a non-uniform charge density lambda = Ax, where x is measured from the center of the rod and A is a constant. How to Find the Moment of Inertia of a Non-Uniform Density Thin Rod about a Given Axis Perpendicular to it. Integrate to find the total V.6. Thats going to be the surface area of the sphere times its thickness. It is an outcome of 23-year long toil of Physics expert who has made it a mission to simplify the complexities of Physics.Ashish Arora, the brain behind this interactive unique website, has all his lectures available on web for free of cost. A charge of uniform linear density 2.0nC/m is distributed along a long, thin, non-conducting rod. Since we will be the same distance from the charge, as long as we are on the surface of this Gaussian sphere, the magnitude of the electric field will be constant over that surface, so we are able to take it outside of the integral and the right-hand side will be, again, q-enclosed over 0. A rod of length $L$ lies along the x axis with its left end at the origin. Q therefore becomes equal to s 4 over R times integral of r3 dr. As we add all these incremental, spherical shells to one another, throughout the volume of the whole distribution, the associated radii of these shells will vary from 0, starting from the innermost one, and going out to the outermost one, from 0 to big R. Q is going to be equal to s time 4 over R, integral of r3 is r4 over 4 which will be evaluated at 0 and big R. Here, we can cancel this 4 in numerator with the one in the denominator, leaving us Q is equal to s, and substituting big R for the little r we will have R4 over here, R in the denominator, and we also have in the numerator. Again, it is in radial direction, so we can express this in vector form like this. Now, we cannot do that because the charge density is not constant. It. Therefore we take the integral of both sides, so we will end up with the total charge on the left-hand side. If we express then E, which will be equal to s divided by 40R. If you call that one Q, q-enclosed is going to be equal to big Q. Lets assume that our point of interest, P, is somewhere over here. Let's say, with length, L, and charge, Q, along it's axis. i know that you have to integrate dq over the rod, what I'm saying is what do you integrate dv over? Charge is distributed non-uniformly throughout the volume of the distribution, which has radius of big R, and the charge density was given as a constant s times little r over big R, and little r is the location of the point of interest. So we will have E times integral of dA over the Gaussian surface s, since cosine of 0 is 1. If you choose one of these shells at an arbitrary location inside of the distribution, something like this, that it has a very, very small thickness, a shell, and it has the radius of r and thickness of dr, and this dr is so small such that when we go from inner surface to the outer surface of this spherical shell, the change in density is negligible. Now we will go back to our Gausss law expression and substitute this for q-enclosed. What wed like to do is calculate the electric field of such a distribution at different regions. (a) 0.984 V (b) 1.37 V (c) 6.72 V (d) 2.31 V 1 It says to use the variables as necessary. As a matter of fact, when little r becomes big R at the surface of the distribution, then it reaches to its maximum value, which is going to be equal to this constant, s coulombs per meter cubed. This R4 and that R will cancel, leaving us R3 in the numerator, so the net charge of this distribution is going to be equal to sR3. Calculate: a. Naturally, it will have the radius of little r. Gausss law states that E dot dA integrated over this surface s is equal to net charge enclosed inside of the region surrounded by this Gaussian sphere divided by 0. Non-uniform lateral profile of two-dimensional electron gas charge density in type III nitride HEMT devices using ion implantation through gray scale mask. Example 2: Potential of an electric dipole, Example 3: Potential of a ring charge distribution, Example 4: Potential of a disc charge distribution, 4.3 Calculating potential from electric field, 4.4 Calculating electric field from potential, Example 1: Calculating electric field of a disc charge from its potential, Example 2: Calculating electric field of a ring charge from its potential, 4.5 Potential Energy of System of Point Charges, 5.03 Procedure for calculating capacitance, Demonstration: Energy Stored in a Capacitor, Chapter 06: Electric Current and Resistance, 6.06 Calculating Resistance from Resistivity, 6.08 Temperature Dependence of Resistivity, 6.11 Connection of Resistances: Series and Parallel, Example: Connection of Resistances: Series and Parallel, 6.13 Potential difference between two points in a circuit, Example: Magnetic field of a current loop, Example: Magnetic field of an infinitine, straight current carrying wire, Example: Infinite, straight current carrying wire, Example: Magnetic field of a coaxial cable, Example: Magnetic field of a perfect solenoid, Example: Magnetic field profile of a cylindrical wire, 8.2 Motion of a charged particle in an external magnetic field, 8.3 Current carrying wire in an external magnetic field, 9.1 Magnetic Flux, Fradays Law and Lenz Law, 9.9 Energy Stored in Magnetic Field and Energy Density, 9.12 Maxwells Equations, Differential Form. Break the total charge into infinitesimal pieces dq.3. (You may be able to determine the answer without the integral. 32. dV is the contribution from dq to the potential at A (with respect to infinity). In other words, we will apply the same procedure that we did in the previous part, except instead of integrating and adding the dqs from 0 to big R, now we are going to add them from 0 to little r. 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