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 .