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].

First, the potential of a ring of charge with a uniform linear charge density and radius R was found to be equal to: 

 (Subscript[V, ring]) = ((2 R k π λ)/(Sqrt[(R^2)+(z^2)]));      (k = (1/(4 π ϵo)))
But the equivalent line charge density is λ = ((Subscript[Q, ring])/(2 π R))    
which ⇒ (Subscript[V, ring]) = ((2 R k π  ((Subscript[Q, ring])/(2 π R)) )/(Sqrt[(R^2)+(z^2)])) = ((k (Subscript[Q, ring])  )/(Sqrt[(R^2)+(z^2)])) .   

For our ring:   (Subscript[Q, ring]) = (Subscript[Q, ring of radius r and thickness dr])= 2 π r dr σ;   (you could call this dQring, a differential of charge, if you wished)
	ding-dong:  note that here we use r, not R because we are going to integrate over r.

(Subscript[V, ring]) turns into a differential, (Subscript[dV, disk]):
The differential of the potential due to this ring of radius r  =  (Subscript[dV, disk])[z] = ((k 2 π r dr σ )/(Sqrt[(r^2)+(z^2)])).
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 Shift-Return  (Shift-Enter)

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 (Subscript[dV, disk])[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 -zo to + zo. 

I chose k =1, R=1, and σ=1 and I set zo = 10.  [I suggest you use a σ >0 to help interpret your result.]

(* 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:

		(Subscript[Q, disk]) = (Area of Disk) * σ = π (R^2) σ

Here we plot (Subscript[V, disk])[z] and the potential of a point charge:  ((k (Subscript[Q, disk]))/(Abs[z]))  (We insert the Abs[z] to replace z because for +(Subscript[Q, disk]), V of a point charge is positive for both +z and -z;  Abs[z] takes care of this).

Qdisk = π  (R^2) 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 zo, over the range -zo to + zo.

As before, I used k =1,  σ=1, R = 1;   I set zo = 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, (Subscript[E, disk]).   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 zo where the z component is the only non-zero component (reasonable:  Vdisk[z] depends only on z;  therefore, at the point P, (OverVector[Edisk])[z] = Ediskz[z] (OverHat[z])

Using Vdisk[z] for the potential we get an If statement in the resulting (OverVector[Edisk])[z].  Looking carefully, we can conclude that (OverVector[Edisk])[z] for z < 0 equals (-) (OverVector[Edisk])[z] for z > 0.     [Hopefully you agree that:   (OverVector[Edisk])[z] = Ediskz[z] (OverHat[z]) .]

(i) Now Plot Ediskz[z] for the same k, σ, R and for the same zo, i.e., over the same range -zo to + zo you used above.  [You will need to grab the z component of  (OverVector[Edisk])[z], which we call Ediskz[z].]

(*  Input Cell *)
Ediskz[z_] =   Edisk[z]  [[3]]  (*  quick way to pull out the z component of (OverVector[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["(Subscript[E, disk])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:)



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 (Subscript[V, disk]) and (Subscript[(OverVector[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: (Subscript[V, disk])[x,y,z] depends in general on x,y, AND z.

Button["1  I agree.  Off the symmetry axis, (Subscript[V, disk]) and (Subscript[E, disk]) 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, (Subscript[V, disk]) and (Subscript[EE, disk]) 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.