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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 84 "                       \+
                                Electrostatics - Electroscope" }}
{PARA 256 "" 0 "" {TEXT -1 540 "   In this maple worksheet, we look at
 the simple electroscope - two like charges suspended from strings.  A
ssume we know the mass of the bobs, and the length of the strings.  Th
en, if we know the angle between the strings, we can easily figure out
 the magnitude of the charges on the bobs.  But the reverse is not so \+
easy!  If we know the charges, it is a very complicated equation to ge
t the angle.  (It can be approximated for small angles, but the exact \+
solution is too complicated to do by hand!)  [Also, see the Physlet EL
ECTROSCOPE!]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 560 "The problem is this : two equal masses, with equal and like ch
arges are suspended from strings from a common point.  The length of e
ach string is L.  The charges will feel a tension force up along the s
tring, gravity pulling down, and the electrostatic force sideways (awa
y from the other charge).  The angle Theta is measured from the vertic
al.  Looking at one of the charges, the vertical component of the tens
ion is equal to the weight of the mass, and the horizontal component o
f the tension is equal to the electrostatic force between the charges.
  Thus  " }{TEXT 259 14 "Tcos(theta)=mg" }{TEXT -1 8 "   and  " }
{TEXT 260 22 "Tsin(theta)= kq1q2/r^2" }{TEXT -1 9 "   where " }{TEXT 
261 14 "r=2Lsin(theta)" }{TEXT -1 58 ".  Eliminating T from these two \+
equations, we arrive at : " }{TEXT 258 44 "tan(theta)sin^2(theta) = k*
q1*q2/(4*m*g*L^2)" }{TEXT -1 399 ".   If we know the angle, it is easy
 to find the magnitude of the charges, but the reverse is not true.  T
o solve for theta, knowing the charge, there is no easy exact analytic
al solution, but maple can find a very complicated one!  Also, we note
 that for small angles, we could  approximate the tan(theta) and the s
in(theta) as being just theta .. and we could get an approximate solut
ion of  :    " }}{PARA 0 "" 0 "" {TEXT 257 34 "     theta^3 = k*q1*q2/
(4*m*g*L^2)" }{TEXT -1 193 ".  The following worksheet will illustrate
 the general (complicated) solution, and allow you to plug in numbers \+
for the various constants .. and then will illustrate the approximate \+
solution.  " }{TEXT 262 128 "Note: this approximation is pretty good .
. up to about 40 degrees, there seems to be very good agreement betwee
n the two values!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 79 "restart;a:=k*q0*q0*(1E-12)/(4*mass*g*(L/100)
^2):  # these are all the constants" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 291 "This next line will generate the solution to the exact equatio
n.  It is very long and complicated (and relatively \"meaningless\" to
 us!).  If you want to supress it from showing up on the screen, chang
e the character at the end of the next line from \";\" to \":\" - that
 will supress the output." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
59 "mytheta :=solve(tan(theta)*sin(theta)*sin(theta)=a,theta); " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "L:=20: mass:=0.5: q0:=2: g:=
9.8: k:=9E+9:   # Change these values if you wish" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 127 "x:=evalf(mytheta): \"This is the exact solu
tion :\"; evalf((180.0/Pi)*x[1]);  # this gives the angle in degrees f
or your numbers." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 124 "This next li
ne is the \"small angle approximation\" - note how close it comes to t
he exact equation .. up to about 40 degrees!" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 114 "mytheta_approx :=solve(theta^3=a,theta): \"This
 is the approximate solution :\";evalf((180.0/Pi)*mytheta_approx[3]);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "10" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 
1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }