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