Solution to Potential and E-Field of Charged Disk ALONG Symmetry Axis

Solution to Potential and E-Field of Charged Disk ALONG Symmetry Axis Problem: Consider a disk of radius R with a uniform charge density σ. Find the Electric Field due to this charge distribution on the axis of symmetry (z axis) for both z > 0 and z < 0. Denote the distance along the z axis from the center of the disk (O) to the point P (on the z axis) by z. A couple of reminders: 1. (* This is a comment *) and 2. If you get some weird results at some point, try going back and re-executing all the previous cells to reset your functions and variables. (a) Start by finding a differential of electric potential dVring[z] (dVring as a function of z) at the point P due to a ring of charge of a radius between r and r + dr. Then integrate over r to find Vdisk[z]. 2 First, the potential of a ring of charge with a uniform linear charge density and radius R was found to be equal to: Vring = 2Rkπλ R2 +z2 ; (k = 1 ) 4 π ϵo But the equivalent line charge density is λ = which ⇒ Vring = 2Rkπ For our ring: Qring Qring 2πR R 2 +z 2 = k Qring R2 +z2 Qring 2π R . = Qring of radius r and thickness dr = 2 π r dr σ; (you could call this dQring, a differen- tial of charge, if you wished) ding-dong: note that here we use r, not R because we are going to integrate over r. Vring turns into a differential, dVdisk : The differential of the potential due to this ring of radius r = dVdisk [z] = k 2 π r dr σ . r2 +z2 We integrate over r from 0 to R to get Vdisk[z]. (b) Enter your equation for dVring[z] in an appropriate integral written in Mathematica and thereby find the equation for Vdisk[z]. A series of assumptions (more than you need) are provided that make the integration run like a champ. They start with: $Assumptions = ...... Reminder: To execute each cell, click your mouse anywhere inside the cell and then hit ShiftReturn (Shift-Enter) 3 Below is an input cell you can use for finding V[z] : (* Input Cell *) ClearAll["`*"] $Assumptions = R > 0 && R ∈ Reals && z ∈ Reals && z ≠ 0 && k ∈ Reals && k > 0 && σ ∈ Reals && σ > 0; (* ENTER in this cell (below this comment) your dVdisk [z] into the appropiate integral --reminder: Limits needed *) You should now have the scalar function Vdisk[z] defined as an algebraic expression which contains the parameters k, R, λ, and the variable z. (c) Assume some values for k, R, and σ; plot Vdisk[z] from some -z I chose k =1, R=1, and σ=1 and I set z result.] o o to + z o . = 10. [I suggest you use a σ >0 to help interpret your (* Input Cell -- enter your code and evaluate*) (d) Clear[k,σ, R] and then evaluate Vdisk[z] to see that it looks ok. (I’ll do it for you): Clear[k, σ, R] Vdisk[z] (e) Interpret the plot of Vdisk[z] you obtained. Enter your Discussion (this is a text cell): (FYI- extra info) You should note that in both directions, the potential falls to very small values for z >> R; the disk starts looking like a point charge. The total charge on the disk is: Qdisk = (Area of Disk) * σ = π R2 σ Here we plot Vdisk [z] and the potential of a point charge: k Qdisk Abs[z] (We insert the Abs[z] to replace z because for +Qdisk , V of a point charge is positive for both +z and -z; Abs[z] takes care of this). 4 Qdisk = π R2 σ; k = 1; σ = 1; R = 1; PlotVdisk[z], k Qdisk Abs[z] , {z, - 20, 20}, PlotRange → {0, 10} As expected, at sufficiently large |z|, the two potentials (the disk and the point charge) are indistinguishable; both go to zero for z -> ∞. (f) So let’s handshake on the presence of a term in Vdisk[z] containing Abs[z]. To find the E field from the potential M or we must take derivatives. M does not like taking the derivative of Abs[z] with respect to z. We make life a lot easier for M and for ourselves if we divide the solution into two parts for z>0 and z<0. We can combine them with an If statement. It will look like this: Vdisk[z_] = If[z < 0, (Stick in here your Vdisk[z] for z < 0), If[z > 0, (Stick in here your Vdisk[z] for z > 0)]]. M is happy taking derivatives of Vdisk[z] in the form of the If statement; it simply performs it for each part separately. Thinking carefully, determine the appropriate Vdisk[z] for the two signs of z and enter into the input cell below your resulting Vdisk[z] [in the form of an If statement]. Execute the cell so M has Vdisk[z] defined. Stick in a Clear[k, σ, R] to keep things honest. (* Input Cell - write your Vdisk[z] for both signs of z; you can use an If statement OR the Piecewise function *) Clear[k, σ, R] (g) For comparison with the plot of Vdisk[z] above, Plot this new Vdisk[z] for the same k, σ, R and for the same z o , over the range -z o to + z o . As before, I used k =1, σ=1, R = 1; I set z o = 10. (* Input Cell *) The plots for the two forms of Vdisk[z] are the same, i.e., for both ± z; the SIGN of z automatically handled this for the first version. (h) Now find the Electric Field, Edisk . I used M’s Grad function in Cartesian Coordinates (which generates a VECTOR). 5 (h) Now find the Electric Field, Edisk . I used M’s Grad function in Cartesian Coordinates (which generates a VECTOR). Clear[k, σ, R]; Vdisk[z] (* just to check to see it is still defined *) This generates a 3D vector (for both signs of z o where the z component is the only non-zero component / (reasonable: Vdisk[z] depends only on z; therefore, at the point P, Edisk[z] = Ediskz[z] z Using Vdisk[z] for the potential we get an If statement in the resulting Edisk[z]. Looking carefully, we can conclude that Edisk[z] for z < 0 equals (-) Edisk[z] for z > 0. [Hopefully you agree that: / Edisk[z] = Ediskz[z] z .] (i) Now Plot Ediskz[z] for the same k, σ, R and for the same z o , i.e., over the same range -z o to + z you used above. [You will need to grab the z component of Edisk[z], which we call Ediskz[z].] Input Cell *) (* Ediskz[z_] = Edisk[z] [[3]] quick way to pull out the z component of Edisk[z] *) (* k = 1; σ = 1; R = 1; (* need these defined for the Plot *) Plot[Ediskz[z], {z, - 10, 10}, PlotRange → {- 7, 7}, AxesLabel → {"z", "Ediskz[z]"}, PlotLabel → Style[Framed["Edisk z[z]"], 16, Purple, Bold, Background → Lighter[Green]]] (j) Write down a brief interpretation/discussion about the two plots (e.g., SIGNS and the sign of λ). How are these two plots are related (hint: SLOPE of one of them)?? Enter your Discussion (this is a text cell): Interpretation: First, we compare this curve with the above Vdisk[z], (I’ve copied and pasted these plots from above to help you get started:) 1 + z2 -Abs[z]) Vdisk [z] = 2 π ( V[z] 6 5 4 3 2 1 -10 -5 0 Edisk z[z] 5 10 z o Enter your Discussion (this is a text cell): 6 Interpretation: First, we compare this curve with the above Vdisk[z], (I’ve copied and pasted these plots from above to help you get started:) 1 + z2 -Abs[z]) Vdisk [z] = 2 π ( V[z] 6 5 4 3 2 1 -10 -5 0 5 10 5 10 z Edisk z[z] Ediskz[z] 6 4 2 -10 -5 z -2 -4 -6 Your Turn: (k) No Brainer - Click inside the cell below (or select it by clicking on the bracket to the right) and execute it (Shift-Return); Answer Boxes will appear; Click on the one you think is correct answer for this question:  Question: Off this symmetry (z) axis, I expect Vdisk and E disk to depend on z only. [Live it up! Click both.] Imagine moving a +q test charge around the disk with uniform + σ at various x,y,z values off the z axis. I think you can see that the off axis solution: Vdisk [ x, y, z] depends in general on x,y, AND z. 7 Button[ "1 I agree. " Off the symmetry axis, Vdisk and Edisk depend on z only", {Print[ Wrong --The symmetry of the problem is broken: in Cartesian Coordinates, we therefore expect x and/or y dependence to creep in. "]}] Button["2 I disagree; Off the symmetry axis, Vdisk and EEdisk generally do not depend on z only ", {Print[" Correct -- The symmetry of the problem is broken; in Cartesian Coordinates, we therefore expect x and/or y dependence to creep in.\n\nIn Spherical Coordinates one would expect θ dependence in V and E, but no ϕ dependence."]}] So No. 2 is the correct answer.