A large brass washer has a 2-cm inside diameter, a 5-cm outside diameter, and is

Question

A large brass washer has a 2-cm inside diameter, a 5-cm outside diameter, and is 0.5 cm thick. Its conductivity is Sigma = 1.5 x 10^7 S/m. The washer is cut in half along a diameter, and a voltage is applied between the two rectangular faces of one part. The resultant electric field in the interior of the half-washer is E = (0.5/rho)a_Phi V/m in cylindrical coordinates, where the z axis is the axis of the washer, (a) What potential difference exists between the two rectangular faces? (b) What total current is flowing? (c) What is the resistance between the two faces?

Explanation / Answer

Conductivity = (1/resistivity)

So, Resistivity = (1/conductivity)= (1)/1.5*107 = 0.66 *10-7 ohm. Meter

Resistivity, = 0.66 *10-7 .m

Area of rectangular face = length * breadth

Here length means difference between inner and outer radius = 1.5 cm= 1.5 *10-2 metre

So, Length of rectangle = 1.5 cm

Breadth = thickness = 0.5 cm (Given)= 0.5 *10-2 metre

Therefore,

Area , A = (1.5 *10-2)*( 0.5 *10-2) = 0.75 *10-4 m2

Total length L = 5*10-2 m (diameter of outer side)

By the definition of electric field,

E = V/L

So,

V = E*L = (0.5/)* 0.5 *10-2

= (0.5/(0.66 *10-7))* 0.5 *10-2

= (0.7575 *107))* 0.5 *10-2

= 1.515 *10 9 volt

Now, E can be defined as,

E = .J =

V/L = .J

J = (V/L)*(1/ ) = E/ = (0.5/)/ = 0.5/2 = 0.5/(0.66 *10-7)2 = (0.5/0.4354)*1014 =1.148 *1014 A/m2

But, J= current density and it is,

J = (I/A)

Therefore,

I = J*A

= (1.148 *1014)*( 0.75 *10-4)

= 0.861 *1010 ampere

Now we know the value of V and I, so we can calculate the value of resistance.

By Ohm’s Law,

V = IR

Therefore,

R = V/I

= (1.515 *10 9)/( 0.861 *1010)

= 1.759 *10-1

= 0.176 ohm

So answers are,

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