Download PH 316 Worksheet MJM September 6, 2005 - Rose

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PH 316 Worksheet
MJM
September 2006
Name_____________________ Box ______
For a circular ring of radius a, and charge
density  C/m, the electric field on the axis
of the ring at a distance z from the center
of the ring is as we worked it out in class:
z
Ering = k(2a)z/(z2 + a2)3/2 = kQz/(z2 + a2)3/2 .
Use this result to find the formula for E on the
axis of a solid disc of radius a whose charge
density is  C/m2, on the axis of the disc at a
distance of z from the disc's center.
You will need to integrate the E fields of a series
of infinitesimal rings. The formula above applies,
but Q refers only to the infinitesimal charge dQ on
one ring.
For the line charge on p. 63 we had dQ = dx,
where  was the charge per unit length. Now,
for the disc, you will have dQ = dA, where
dA is the area of an infinitesimal ring of radius a
and thickness da, and  is the charge per unit area.
You must work out the infinitesimal area dA and do the integration in closed form.
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Now you are to find the electric field at the center of a hemispherical bowl which is uniformly
charged, having a surface charge density of  C/m2 . You must again work out the infinitesimal area
element dA, but this time the integration variable is not the radius a
Finally, you are to find the electric field at the center of a hemisphere of uniform charge density 
C/m3. This involves integrating over the volume, using dQ = dV. This involves the integration over
two variables, in order to fill the volume of the sphere. You will get a finite result for this integral even
though some of the charges are infinitesimally close to the center, and should give huge E fields.