Complete Solution to Potential and E-Field of a sphere of radius R and a charge density ρ[r] = CC r 2 and r n

Transcription

1 Complete Solution to Potential and E-Field of a sphee of adius R and a chage density ρ[] = CC 2 and n Deive the electic field and electic potential both inside and outside of a sphee of adius R with a adially symmetic chage distibution given by ρ[] = CC 2. Use M as much as possible - some deivation by hand is usually needed. (a) Fist, fo ρ[] = CC 2, detemine the total chage Qinside[] contained within the symmetically placed sphee of adius R In[61]:= (* Input cell *) CleaAll["`*"] ρ[_] = CC 2 ; Out[63]= Qinside[_] = ρ[] π pime 2 dpime 0 CC π 5 3 (b) Find the total chage Qtotal contained in the entie sphee: We find Qtotal by evaluating Qinside at = R: In[6]:= (* Input cell *) Qtotal = Qinside[R] Out[6]= 3 CC π R5 (c) Detemine the E field EE[] fo all ; Use the above Q functions and take advantage of the spheical symmety. (This is a text cell -- Type in you deivation)

2 2 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb We use Gauss's Law. Fo spheical symmety whee ρ depends only on, Gauss's Law simplifies to the following: (i) Outside, the E-field at >= R will look like that of a point chage with magnitude Qtotal. (ii) Inside, the E-field at <= R will be dependent only on the chage inside a sphee of adius, i.e., Qinside[]. This Q will look like a point chage at the oigin with magnitude Qinside[]. Both fields will be adial. these statements: If not convinced go to this site; it will einfoce ou undestanding of They also discuss in detail the case whee the chage is distibuted unifomly thoughout the sphee, i.e., ρ is a constant, i.e., popotional to o = 1). (i) implies OUTSIDE: EEoutside[] = k Qtotal " 2 (ii) implies INSIDE: EEinside[] = k Qinside[] " 1 whee k = 2 π ϵ o (d) Ente expessions in an input cell fo the magnitudes of the E fields inside and outside the sphee using Qtotal and Qinside[]; these ae functions that you found above. Note: M emembes the Qtotal and Qinside[] functions once the cells defining them (above) ae executed. In[65]:= (* Input cell *) Out[65]= Out[66]= EEoutside[_] = EEinside[_] = CC k π R CC k π 3 3 k Qtotal 2 (* magnitude only *) k Qinside[] 2 (* magnitude only *) (e and f) Use the above E-fields to find Voutside[] and Vinside[]. (I enteed a bunch of assumptions to make M happy. BTW: I m being sloppy about things like > R vs. >= R. To M: No Big Deal. I usually, but not always, stick in the = sign). (e) Fist, Voutside[], >= R We stat with V[] - V[ ] = - E[pime] dpime, whee pime is a dummy vaiable. [this expession is the wok pe unit chage to move a test chage fom infinity to the position. Just a

3 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb 3 expession is the wok pe unit chage to move a test chage fom infinity to the position. Just a eminde -- this all holds fo adial dependent functions. NOTE - sign. If you ae toubled by this, go hee: http : // ~acosta/ phy2061/ lectues/ ElecticPotential.pdf Theefoe, to find Voutside we fom the following integal (magnitude only) Voutside[] = - EEoutside[pime] dpime = - k Qtotal pime 2 dpime. pime is a dummy vaiable. In[67]:= (* Input cell *) $Assumptions = (R pime ) && pime Reals && Reals && R; Voutside[_] = - EEoutside[pime] dpime (* EEoutside[] is defined above *) Out[68]= CC k π R 5 3 (f) Now find Vinside[] <= R. To find Vinside we integate the Eoutside[] fom to R (which equals Voutside[R]) and then integate Einside[] fom R to ; I call this contibution Vinsideonly[] (of couse we inset the - sign). Add the two togethe. The limits on these integals ae consistent with moving the test chage fom to an inside the sphee, < R. Note: I have changed the assumptions to aid the Vinsideonly[] integation. Recommend using Simplify nea o at end. In[69]:= (* Input cell *) $Assumptions = ( 0 pime R) && pime Reals && ( 0 R) && Reals; Vinsideonly[_] = - EEinside[pime] dpime R Vinside[_] = Voutside[R] + Vinsideonly[]; Vinside[_] = % // Simplify Out[70]= - 3 CC k π - R Out[72]= CC k π - 5 R (g) Put the two togethe with an If statement like this: VV[_] = If[ R, Voutside[], If[ R, Vinside[]]], which defines the potential fo all. In[73]:= (* Input cell Done fo you: *) VV[_] = If[ R, Voutside[], If[ R, Vinside[]]];

4 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb (h) Do the same fo E: call it EE[_] (using anothe If statement) In[7]:= (* Input cell Done fo you: *) EE[_] = If[ R, EEoutside[], If[ R, EEinside[]]]; (i) Plot VV[] and EE[] togethe. (I used Show) Don t foget you have to give values fo k, CC, and R fo M to plot. I used: k = 1; CC = 1; R = 1; In[75]:= (* Input cell *) k = 1; CC = 1; R = 1; pp = Plot[{EE[], VV[]}, {, 0, 6}, PlotRange {{0, 6}, {0, 6}}, PlotLegends {"EE[]", "V[]"}]; line = Gaphics[Line[{{R, 0}, {R, 6}}]]; Show[pp, line, PlotRange {{0, 6}, {0, 6}}] (* The vetical line shows the edge of the sphee, = R *) 6 5 Out[78]= 3 2 EE[] V[] (j) I d now like to exploe V and E fo ρ[] = CC n whee we can vay n. CHARGE DISTRIBUTION ρ[]! Reminde: this is the n in the I ve combined all the cells used above and have done the heavy lifting fo you on the Manipulate which allows you to see what happens when you change n. READ the Pint Outputs caefully to undestand the equations showing up in the Output. Play with n and examine the shape of the esulting V and E functions. A vetical line indicates whee = R. If you look at the equations fo Vinside, you will see that cetain values of n lead to poblems (ty to figue out what they ae). I have tied to limit the ange of so the plots behave.

5 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb 5 figue out what they ae). I have tied to limit the ange of so the plots behave. Below you will be asked to do some intepetation. In[79]:= (* Execute this whole cell *) CleaAll["`*"] ρ[_] = CC n ; Pint["ρ[] = ", ρ[]] $Assumptions = 0 R && Reals && R > 0 && R Reals && n > -3; Pint[] Qinside[_] = ρ[pime] π pime 2 dpime; 0 Pint["Qinside[] = ", Qinside[]] Qtotal = Qinside[R]; Pint["Qtotal = Qinside[R] = ", Qtotal] EEinside[_] = k Qinside[] 2 ; Pint["EEinside[] = ", EEinside[]] EEoutside[_] = k Qtotal 2 ; Pint["EEoutside[] = ", EEoutside[]] $Assumptions = R && Reals && R > 0 && R Reals && n > -3; VVoutside[_] = - EEoutside[pime] dpime // Simplify; Pint["VVoutside[] = ", VVoutside[]] VVoutside[R]; $Assumptions = 0 R && Reals && R > 0 && R Reals; VVinside[_] = VVoutside[R] - EEinside[pime] dpime; R VVinside[_] = VVinside[] // Simplify; Pint["VVinside[] = ", VVinside[]] EE[_] = If[ > R, EEoutside[], If[ R, EEinside[]]]; VVV[_] = If[ > R, VVoutside[], If[ R, VVinside[]]]; k = 1; CC = 1; R = 1; max = 5; Pint[] Pint["Note all of these functions depend on n!"] Pint[] Pint[ "Hee we Manipulate n to show the behavio of the magnitudes of E and V. I also show ρ[]. The line shows = R. Move the slide to change n (I stat with n = 5; moving the slide left to ight educes n fom 5 to ~ You can also hit the + and - buttons to change n by "]

6 6 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb π Manipulate Show Plot If > R,0, If R, n, If > R,, (3 + n) 2 If R, π 1+n 3 + n, If > R, π π 3 + n - 2+n (3 + n), If R,, {, 0, max}, (2 + n) (3 + n) PlotRange {{0, max}, {0, 15}}, PlotLegends {"ρ[]", "EE[]", "V[]"}, Gaphics[Line[{{R, 0}, {R, 15}}]], {n, 5, -1.25,.2, Appeaance "Open"} ρ[] = CC n Qinside[] = CC π 3+n 3 + n Qtotal = Qinside[R] = CC π R3+n 3 + n EEinside[] = CC k π 1+n 3 + n EEoutside[] = CC k π R3+n (3 + n) 2 VVoutside[] = VVinside[] = CC k π R3+n (3 + n) CC k π - 2+n + (3 + n) R 2+n (2 + n) (3 + n) Note all of these functions depend on n! Hee we Manipulate n to show the behavio of the magnitudes of E and V. I also show ρ[]. The line shows = R. Move the slide to change n (I stat with n = 5; moving the slide left to ight educes n fom 5 to ~ You can also hit the + and - buttons to change n by

7 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb 7 n Out[107]= ρ[] EE[] V[] (k) Intepet the behavio of V and E when you vay n. Look caefully at the plots above as you vay n. Some suggestions fo discussion: What about V and E fo > R and all n? Look at the behavio of E() with espect to the behavio of V (think E = - Gad[V]). Fo < R (inside the sphee): Look at n = 2 (compae with above); n = 1; n = 0 (meaning?); 0 < n < -1; n = -1. Add anything else if you wish. Text Cell : type in you intepetation(s) hee. V and E fo > R and vaious n: Fom Gauss s Law we concluded that OUTSIDE, E and theefoe V looks like that of a point chage at the oigin and magnitude Qtotal. Thus, V ~ 1 and E ~ 1. So they do not change shape 2 with n. (amplitude changes because of n dependence in the Q s) In Geneal: Look at the behavio of E() with espect to the behavio of V (think E = - Gad[V]).

8 8 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb Fo a given R, the slope of V (sign and magnitude) detemines the value of E. Note that fo the n ange chosen, at = R, dv has a maximum negative value, then continues to be negative but d deceasing. Putting in the - sign ( hee E = - dv d ), E should incease with while < R and then decease with outside of the sphee ( > R). Obviously, E has a maximum at = R. Although not obvious at - values of n (but vey convincing at + values), the slope of V at = 0 equals zeo SO E = 0. Look at n = 2 (compae with ou fist esults above). V and E look exactly the same as above (as it should). Inside: VVinside[] = CC k π -2+n +(3+n) R 2+n ; n = 2 -> VVinside[] = (2+n) (3+n) < < R=1, VVinside is a max at = 0 and dops with cuvatue to = R (= 1); CC k π - + R 20. I.e., fo 0 CC k π 1+n EEinside[] = ; n = 2 -> EEinside[] = 3+n CC k π 5 3 I.e., 0 < < R=1, EEinside is zeo at = 0 and ises as 3 to = R (=1) n = 1. Outside > R the cuves that ae consistent with V~ 1 and E ~ 1 2 as expected. Inside: VVinside[] = CC k π -2+n +(3+n) R 2+n ; n = 1 -> VVinside[] = (2+n) (3+n) < < R=1, VVinside is a max at = 0 and dops with cuvatue to = R (= 1); CC k π R I.e., fo 0 CC k π 1+n EEinside[] = ; n = 1 -> EEinside[] = CC k π 3+n 2 I.e., 0 < < R=1, EEinside is zeo at = 0 and ises as 2 to = R (=1) n = 0 (meaning?) This epesents a chage distibution ρ that is constant (unifom). This leads to a linealy inceasing E (pehaps you have seen this in a textbook example). [See again: ] 0 < n < -1; Note that E is cuved with a deceasing ate (magnitude of E with inceasing ). This can be seen by examining the dependence of E fo n < 0. BUT at n = -1: Inside: VVinside[] = CC k π -2+n +(3+n) R 2+n ; n = -1 -> VVinside[] = (2+n) (3+n) < < R=1, VVinside is a max at = 0 and dops linealy to = R (=1) CC k π (- + R). I.e., fo 0 12 EEinside[] = CONSTANT. CC k π 1+n ; n = 1 -> EEinside[] = 3+n 5 CC k π 0 I.e., 0 < < R=1, EEinside is

9 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb 9 CONSTANT. -1 < n < 2; Note the explosive behavio of E and V fo cetain values of n. The most negative n I chose fo the Manipulate was my attempt to keep the plot of the potential inteesting. Hee is V and E fo n = -.8 (you can compae with the Manipulate esults above by setting n = -0.8) Hee ae the functions fo this paticula n (-0.8): In[108]:= Clea[k, CC, R] n = -.8; (* Reminde: this is the n in the CHARGE DISTRIBUTION ρ[] *) VVinside[] EEinside[] Out[110]= CC k R 1.2 Out[111]= CC k 0.2 The minus sign in the tem in VV might bothe you; the R tem wins. Hee is a e-plot of VVinside and EEinside (only coect fo < R) to show that all is well:

10 10 MODULE-SOLUTION-ABSOLUTE-FINAL-ho-CC-2-AND-n-ChagedSphee-V-and-E-vs--READY-FOR-WEB-PAGE-UPLOAD-1-1.nb In[112]:= CC = 1; k = 1; R = 1; n = -.8; EEinside[] VVinside[] Plot[{EEinside[], VVinside[]}, {, 0, R}, PlotRange {{0, 5}, {0, All}}, PlotLegends {"Einside[]", "Vinside[]"}] Out[113]= Out[11]= Out[115]= 6 Einside[] Vinside[] The E and V functions only hold fo < R so the plot is teminated at = R. If you compae with the plot above, you see that they ae the same. Fo n < -2, the infinities cause so many poblems that we close ou eyes and walk away. If you have patience galoe, feel fee to exploe. (hyme unintended).