page 3 of 820A metal sphere of radius A, has a total charge, Q. (Do Not use Gauss’s Law to solve this problem)2) Electrostatic Potential:Objective: To determine inside the sphere using the convolution integral.fV( )r

An arbitrary field point is taken on the z-axis,fˆz(AzA)rza) Write an expression for the surface charge density. ( no derivation is necessary)b) Using the law of cosines, write an expression for . Show all relevant symbols on the diagram.2sfrc) Evaluate the integral for V(z). (You have seen this integral in many examples and homework.)zsS0sf1V(z)da4r(integral warning)let S (r, , ) r A, 0, 02CABθ 222CAB2ABcos()

page 4 of 8Objective: To create expressions for electrical parameters of the lumped element.Two PEC electrodes are attached to the ends of a curved bar.0A()BAr3) Lumped Element:The bar medium has conductivity, and permittivity, a) Determine an expression for inside the bar, f()Erfbarr(integral warning)b) Determine an expression for the lumped resistance.(integral warning)20barlet (,,z) A B,0,0zT  0B()rTzABV+V-

page 5 of 8c) Determine an expression for the lumped capacitance.d) Calculate a numerical value for the capacitance, given, T = 10 cm, A = 1 cm, B = 11 cm and Δφ = 45o.3) Lumper Element: (continued)(integral warning)

page 7 of 8d) Determine an expression for the current, I, in terms of geometric constants, electrical constants and v1.4) Conductive Fluid: (continued)e) The current, I, can be changed by controlling α .Given, B0 = 1 T, W1 = W2 = 5 cm , L1 = 1 cm, and σ = 10-3 S/m,Calculate the flow velocities, v1 and v2 in units of metres/sec13I 0 when I 0.5 A when 1