Question

A thin metal cylinder of length L and radius R1is coaxial with a thin metal cylinder of length L and a larger radius R2. The space between the two coaxial cylinders is filled with a material that has resistivity rho. The two cylinders are connected to the terminals of a battery with potential difference ΔV, causing current I to flow radially from the inner cylinder to the outer cylinder. Part A Find an expression for the resistance of this device in terms of its dimensions and the resistivity. Express your answer in terms of some or all of the variables R1, R2, L, and rho. R = nothing

Answer 1

The expression for resistance is
`R = (\rho)/(2 \pi L) ln[(R_2)/(R_1) ]`

Step-by-step explanation:

Generally flow of charge at that point is mathematically given as

`J = (I)/(2 \pi r L)`

Where L is length of the cylinder as given the question

The potential difference that is between the cylinders is

`\delta V = -E dr`

Where is the radius

Where E is the electric field that would be experienced at that point which is mathematically represented as

`E = \rho J`

Where is the
`\rho` is the resistivity as given the question

considering the formula for potential difference we have

`\delta V = -[(\rho I)/(2 \pi r L) ]dr`

To get V we integrate both sides

`\int\limits^V_0 {\delta V} \, = \int\limits^(R_2)_(R_1) {(\rho I)/(2 \pi L r) } \, dr`

`V = (\rho I)/(2 \pi L) ln[(R_2)/(R_1) ]`

According to Ohm law

`V= IR`

Now making R the subject we have

`R = (V)/(I)`

Substituting for V

`R = (\rho)/(2 \pi L) ln[(R_2)/(R_1) ]`