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THE COULOMB INTEGRALS
-The electric field @ the point with position vector F
due to a ( point ) charge q' located @ the point w/
position vector F 'is "
pz÷-
- II, z
' ⇒ " =t÷÷=¥¥÷ .
- If there are N charges qi'
,i -
-I
, . . .
,N
,located @ points
with position vectors Fi '
,then the total electric
field @ the point w/ position vector F is the sums
of the field produced by each charge :
hi"
Ii
i¥÷i⇒⇒÷¥÷÷÷÷÷i÷÷i.
ai ¥- In principle ,
this is all of electrostatics.
But finding E
for realistic numbers of point charges is intractable so
we usually replace collections of many charges w/
distributesof charge .
The way the charge is
spread at along a curve,
surface,
or volume is
characterized by a charge density .
- For example , spread a total charge Q along a line
or curve .
The Line charge density X l F ' ) tells you
how much charge is found along a tiny lengthde
'
located @ F'
:
a ← An actual physical object like a wire or
rneadsm-otitota.IT?.I.
4 total charm a
X.
I
:÷÷:÷÷÷÷:*:
- Notice that XCF'
) depends on F'
- that is,
different bits
along the length of the curve may have more or less
charge than the other bits .If we add up all the
charges for each bit along the path we get the
total charge :
←
dqCF'
) -
-de
' XCF ' )
Q =
fpdqtr'
I
=fpde'XLF ' )
"Visit every point
along P & add up
the charges .
- If the charge is spread uniformly along P then X is
a constant & each bit has the same amount of charge
on it : If length :
garstang -
ysaP%%.
notQ = fpdl
'X = X fpdl
'= X L ⇒ X =
I -
charge density .
L
- Be Careful ! Don 't pull charge densities out of integrals if
they are not constant !
- To find the electric field @ a point w/ position vector F
produced by charge spread out along a path P,
we just
add up the contributions from all the little charges dqtr 's
like we would w/ point charges .
←Visit every pt . along P
#Add E produced
A @ E by charge
!'
Ek ) = §¥eod9lF' I IT dares to total
= 4¥ fpdl'
XF ' ) zig#=¥o¥x⇒÷:i .
- This is exactly the same thing we did when we had
N point charges . Except now,
N → as and a point
charge q ! located @ Fi ' becomes an infinitesimal bit
of charge dqtr' ) labeledby its position F '
alongthe path .
The sum over an A number of infinitesimal
quantities becomes an integral .
Etr ) = 4÷qfpdl' XLF ' ) IT
- What if we spread a charge Q over some surface
8 ? The surface charge density Otr ' ) tells us how
much charge is present on an infinitesimal area
da ' located at the point F 'on the surface :
Eiht' thread:'a9 .
charge dqtr' ) = da ' otr ' )
- As with the line charge ,if we add up the charges
on each bit of the surface we get the total
charge Q .
Q = fgdqtr' ) = fgda
' TCF '
)
^ ITT ) The amount of charge per
Visit every pointon surface 5 & bit of area may vary from
add up the bits point - to - point so it dependsof charge
on inhere we are on surface .
- If the charge is spread uniformly over the surface
then the surface charge density is the total chargedivided by the area of the surface :
5 = QIA. In
that case - and on¥ that case - -0 can be pulled at of
integrals .
- The electric field produced at point F by charge spreadout over a surface 5 is
, again , just the sum of the
electric fields produced by all the infinitesimal bits
of charge dqCF' ) spread over the surface .
Visit every
,Add that point 's car -
Ept. on 5 tributes to electric
5 -
ETF ) = ftp.dqcryiafielder .
E- F - F .= ¥eo§da ' FCF ' ) ÷,it= ¥qfgda
' -0K ' )3
- And finally ,we might spread a total charge Q
throughout the interior of some volume V.
The
( volume ) charge density FCF' ) tells vs how much
charge is found in the infinitesimal volume de '
located @ point F 'within the volume .
÷:÷¥÷÷÷:d qtr
' )=dT'gCF'
)
- Adding up all the bits of charge within the
volume gives the total charge :
Q = fudqtr ' ) = Jude'
Str'
)
- If the charge is spread uniformly throughout V
then g is a constant equal to Q divided by
the volume Vi f = QIV. But we will very often
Consider charge densities that vary from point to
point within the volume !
- The electric field @ f is :
ETH - f¥eodar→E.
= ¥eo[dT' Str' )
E -
- F - F '
= ¥qfudT'gCF'
)✓ If - F' 13
- In all these cases, we set up & evaluate the integrals
the same way .
1) Set up cards that make it easy to describe the
Charge distribution
2) Find F ' for points on P,
5,
or V
3) Work out £12 '
4) Assemble the full integrand
5) Set up & evaluate the integral .
- Be careful ! If the charge density varies from point - to -
point you can 't pull it outside the integral .( This is a
Common mistake ) .
- As an example , suppose I have a cylinder w/ height h
& radius R,
and charge density that depends on the
height of the point above the base of the cylinder .
Whats the Coulomb integral for the Electric field @
a point on the Z - axis outside the cylinder ?
A 1) Cylinder : Use CPC w/ OES ' ER,
OE Z' Eh,
•
and O E 0/42 it . Base of cylinder is'
¥÷÷÷÷÷:÷÷:÷÷:÷:÷.
# the function is ! ) Use CPC I cartueeqgysy.itI
2) F' = s 'S t Z' I = s 'cos¢' Its '
sincflytz ' I
F = O x' toy t ZI
3) I = F - I '= - s
'cos ol ' I - s
' singly t L Z - z' ) I
IN = f0'toz = FEET
% = Eg,
e-sicosofx-sis.no/'ytLz-zDE-( s
' '+ C z - z 'T )
" '
4) de'
= s 'd¢' ds 'd 't'
in CPC
↳ de
'gLztz÷=dcfds'dz'get
'
)£s'izI( s' Zt ( z - z c) 2) 312
5) ECO,O,'t ) = y÷q§%§hzi§s ,
gczytskosdx-sks.no/'yts#z-5z)( s
' It I z - zig )312
This integral gives you E @ the point C 0,0 , 't ) .You can 't
give a complete answer w/out knowing glz ),
which I did
not specify .However
, you Lan show that Ex E: Ey are
both Zero @ points on the Z-axis ( but not @ other
points ) .Notice that the only ¢ '
dependence in the integrandis cos of
'in the x - comp . and sin of
'in the y - camp .
Integratingeither of these over ¢ ' from 0 to 2 it gives 0 :
f¥f' cos ¢'
= folio 's .no/'-- O
↳ Elo ,0 , 't ) = z÷eEf§zi§s .
gczi )s'×CZtI
I y
( s' 't Cz - E) 2)
3k
Notice that we integrate over ad the prime coordinates,
So our final answer depends only on Z,
h,
R,
e: the details
of g !
- What about a constant X on a circular wire of
radius R ? What is E @ an arbitrary point I x , y ,z ) ?
Z^
÷ . a
" I:It:"
ins .
So pts on ring have s'
= R,# y
Z' = O,
and OE 0/22 IT .
xL
2) F' = 125+0 I = R cos ¢' I t R sin ¢
'
y'
t O I
F = xxntyytz I
3) I = ( x - Rosol'
) I t ly - Rsincf' ) i t -2 I
IEI = #( X - Rcoscf' ) 't ly - Rand
' ) 'tZZ
§,
=
l x - Rosol'
) I t ly - Rsm 4' ) it 2- I
-
( ( X - Rosol'
) 't ly - Rang )-
t z 2)" 2
4) de'
= Rollo'
,X = canst .
5) EH, y ,Z ) = ¥qJ§ R × lx-Rcosotlxtly-Rs.no/')ytzz(lx-Rcoso/')2tly-Rsino/')2tz2
)" Z
There are some things we could do to simplify this,
but even so its a tough integral . Coulomb integralsare usually very difficult unless the chargedistribution
is highly symmetrical arid we try to
calculateE @ a point that doesn't spoil that
symmetry !
- One last note : We use primed coordinates to label points
along a curve, on a surface
,or in a volume that contain
charge . We integrate over those variables when we
calculate E, so the final result shellnot contain
any primed coordinates in it !
- In the last example E depends on the coordinates F of
the point where we are computing the field, as well
as the radius R of the ring and the constant charge -
per - unit - length X.
But it doe not depend on 4'
! That
refers to a specific point on the ring ,and we visit all
those points in the integral .