Assignment # 4 SolutionsBdE@
Consider the components of 8 and FIR
Tx 2yJ taek ) .
( Erik it Ey Kj + Ez gate )
The first term well yield :
XxFDR t Ex @xk )and each subsequent term well be the same with
x replacedby yand z
. Thus we will have terms
that look like 42 ( I.E,) from the first terms
,
and E,
. ch ) from the second terms. Adding these
together we have
I.
ie) -- ft.IE,)y
,
t I, Iqanaemia
With this we car rewrite Uw as follows .
U ,z=- tofdsr ( I. ⇐yd - E. 4
. )
With thedivergence theorem we can convert
the above into :
§dA . ( Eik )
This corresponds to an integral at r - soo.
Whilethe surface area grows as r2
,the integrand
tends to zero as Yrs .
This the integral as
awhole
goesto zero .
Now,
we can utilize Gauss '
Law andrecognize
that the divergence of Tz,
can be replaced with
the charge distribution associated with E,
.
c. e . I.E,
= ¥.stir - is.
( Note : We can
ignorethe other charge for the
same reason that we are
expressingthe electric field
as the sum of the field created by q ,
and qz . )
Now this integral becomes :
fdsrqsor-ri ' KEITHThis should be familiar !
I
This should bevery
reminiscent of the
pattern shown by a dipole ( or at least half
of the field ) .
Now if we add an additional chargeto create the appropriate field
,our field
must respect the cylindrical symmetryof our
charge distribution .Therefore
, it can only be
either purely perpendicular or purely parallelto the conducting plane . Since the charges are
opposite signed ,we expect it will be purely .
This makes calculating the magnitude easier,
since we only need to
worryabout the I
component .
E.ir#kefa..aEI:Iga-axFaIITeial/*o=-2ak9ff*a--2692 = Ecr )
( aft f) 3k
If we Set a Gaussian surface veryclose
tothe conductor, we
can use the limiting case
to find the charge density .
Eo ( fix ) = ocr ) . DA
to(E - E ) .dA= o Cr ) .dAXZO XEO
to = o or )
Now if we evaluate the integral we find :
JdAoH= - 2keoqaf.gr?Id9-ya= - oh
Finally ,
it is easiest to evaluate the force on
qdue to the fictitious charge a distance Za away .
Fai - Kodi4 AZ