In Lab $2,$ we used a device called Faraday's Ice Pail to measure the

quantity of charge carried by a plastic wand.

For the pail that was used in our lab, the inner basket has a diameter of

$10cm,$ while the outer basket has a diameter of $15cm.$ The thickness of

the wires of the baskets is negligible. The mesh grid formed by the wires

covers approximately $5\%$ of the surface area of each pail.

For simplicity, let us treat the baskets as solid surfaces; treat the height

of the basket as infinite; and treat the charge carried by the wand as

distributed uniformly along center line of the pail $($therefore$,$ the system

now has a cylindrical symmetry$).$

a$)$ Assume the line charge density on the wand is $,$ find the electric

field at $r$ distance away from the wand.

b$)$ When the electric potential of the inner basket is $10V$ lower than that of the outer basket,

calculate the line charge density on the wand. Express your result in $\frac{C}{m}($Coulombs per

meter$).$ What is the sign of the charge?

c$)$ Calculate the capacitance of a $15cm-$tall segment of the ice pail. Express your result in $F$

$($Faraday$).$

d$)$ Calculate the energy stored in the electric field inside the gap between the inner pail and

the outer pail per unit height, when the potential difference between the inner and the outer

pail is $10V.$ Express your result in $\frac{J}{m}.$

e$)$ Now suppose that the wand is no longer centered in the basket $($but still not touching$).$

Would it break the symmetry and change the answer for part $b?$ Briefly explain.

f$)$ Suppose that we transfer all charge from the wand to the inner basket $($so that a segment of

the inner basket with height $h$ will now carry a total charge of $h).$ Will the potential

difference between the inner basket and the outer basket change? Why or why not?

g$)$ Now if we treat the baskets as their actual shape $-$ mesh grids, would you expect the answer

to part e to change significantly? Briefly explain.