A 1.50 m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x from the left end and obeys the formula rho( x ) = a + bx^2, where a and b are constants. At the left end, the resistivity is 2.25 10^-8 Omega*m, while at the right end it is 8.50 x 10^-8 Omega*m.

What is the resistance of this rod?

What is the electric field at its midpoint if it carries a 1.75 A current?

If we cut the rod into two 75.0 cm halves, what is the resistance of each half?

Question

A 1.50 m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x from the left end and obeys the formula rho( x ) = a + bx^2, where a and b are constants. At the left end, the resistivity is 2.25 10^-8 Omega*m, while at the right end it is 8.50 x 10^-8 Omega*m.

What is the resistance of this rod?

What is the electric field at its midpoint if it carries a 1.75 A current?

If we cut the rod into two 75.0 cm halves, what is the resistance of each half?

anonymous · Answer

$$\int\limits_{x=x1}^{x=x2}[a+b^2]dx$$
where a& b can be find by initial conditions =>this give the total resistance. then if u integate it to mid point then u will get the resistance of left side then subtract it from total resistance. this is for hoolow cylinder if it is filled by same material then just multiply the given above equation by area of circle(pi*r^2):)

anonymous · Answer

Electric Field = (current density)/(resistivity)
current density=i/A=>i-current & a = cross sectional area
so, E=i/rho*a:)