Find the resistance of electrons in nanoshells

[Jaccobtw]

Homework Statement

"Gold Nanoshells" are nanoparticles with a spherical glass core and a thin spherical gold shell. They have tunable optical properties, but at the moment consider their electrical properties. If current flows uniformly from the inner surface of the shell to the outer surface, what is the resistance? Assume a typical nanoshell with inner radius 40nm and outer radius 50nm. The resistivity of gold is 2.44∗10 −8 Ωm.

Relevant Equations

$$R = \frac{\rho l}{A}$$

We are given the resistivity of Gold. The length will be the outer radius minus the inner radius. We need to find the area. But I'm not sure how that would work given the spherical nature of the particle Would we need to use an integral to add up the increasing cross sectional area from the inner radius to the outer radius? I'm not sure, what do you think? Thank you

 

[bob012345]

Homework Statement:: "Gold Nanoshells" are nanoparticles with a spherical glass core and a thin spherical gold shell. They have tunable optical properties, but at the moment consider their electrical properties. If current flows uniformly from the inner surface of the shell to the outer surface, what is the resistance? Assume a typical nanoshell with inner radius 40nm and outer radius 50nm. The resistivity of gold is 2.44∗10 −8 Ωm.
Relevant Equations:: $$R = \frac{\rho l}{A}$$

We are given the resistivity of Gold. The length will be the outer radius minus the inner radius. We need to find the area. But I'm not sure how that would work given the spherical nature of the particle Would we need to use an integral to add up the increasing cross sectional area from the inner radius to the outer radius? I'm not sure, what do you think? Thank you

What does your gut tell you?

 

[Jaccobtw]

What does your gut tell you?

I should use the surface area of the inside face of the shell

 

[SammyS]

I should use the surface area of the inside face of the shell

Time for a gut check.

In your OP, you stated:

Would we need to use an integral to add up the increasing cross sectional area from the inner radius to the outer radius?

I thought initially that perhaps that was your gut feeling.

 

[Orodruin]

Your relevant equation holds only for a conductor with constant cross sectional area. Since your conductor does not have constant area, trying to use it is like trying to fit a square peg into a round hole. Knowing and understanding the applicability of different expressions is a fundamental skill in physics.

 

[bob012345]

Would we need to use an integral to add up the increasing cross sectional area from the inner radius to the outer radius?

Yes, do that.

 

[bob012345]

It would be nice to see you , @Jaccobtw , try and work this out so we know you learned something...

 

[Jaccobtw]

It would be nice to see you , @Jaccobtw , try and work this out so we know you learned something...

Here's what I did:

$$4\pi\int_{40 \times 10^{-9}}^{50 \times 10^{-9}} r^2dr = 2.56 \times 10^{-22}m^3$$

This gives the volume from the inner radius to the outer radius but we need essentially the average surface area. So divide this answer by $10 \times 10^{-9}$ to get the average surface area.

Plug in the rest of the numbers:

$$\frac{(2.44\times10^{-8}\Omega m)(10 \times10^{-9}m)}{2.56 \times10^{-14}m^2}$$

And the answer I got was correct. I hope my reasoning was too.

 

[Orodruin]

I do not buy that reasoning, sorry. It might get fairly close to the correct value, but you certainly do not want average area. What you do want to do is to add up all of the small contributions of consecutive thin shells. In essence, for a thin shell of thickness $dr$ (so thin that area change through it is negligible) the small resistance $dR$ in the shell is given by
$$
dR = \frac{\rho \, dr}{A(r)}.
$$
Then you need to sum those contributions up (read: integrate).

 

[Jaccobtw]

I do not buy that reasoning, sorry. It might get fairly close to the correct value, but you certainly do not want average area. What you do want to do is to add up all of the small contributions of consecutive thin shells. In essence, for a thin shell of thickness $dr$ (so thin that area change through it is negligible) the small resistance $dR$ in the shell is given by
$$
dR = \frac{\rho \, dr}{A(r)}.
$$
Then you need to sum those contributions up (read: integrate).

Why do you have an r in the denominator?

 

[SammyS]

Why do you have an r in the denominator?

Actually, that's Area in the denominator. - the area of a spherical shell of radius $r$ .

See your Relevant Equation.

 

[bob012345]

Here's what I did:

$$4\pi\int_{40 \times 10^{-9}}^{50 \times 10^{-9}} r^2dr = 2.56 \times 10^{-22}m^3$$

This gives the volume from the inner radius to the outer radius but we need essentially the average surface area. So divide this answer by $10 \times 10^{-9}$ to get the average surface area.

Plug in the rest of the numbers:

$$\frac{(2.44\times10^{-8}\Omega m)(10 \times10^{-9}m)}{2.56 \times10^{-14}m^2}$$

And the answer I got was correct. I hope my reasoning was too.

The best practice in general is to work out the math symbolically so you can see the relationships between variables first and then put in the numbers.

 

[Jaccobtw]

Actually, that's Area in the denominator. - the area of a spherical shell of radius $r$ .

See your Relevant Equation.

Then I guess that makes sense. Find the resistance through each infinitesimal cross sectional area and add them (the resistances, not the area) up.

 

[Orodruin]

Actually, that's Area in the denominator. - the area of a spherical shell of radius $r$ .

See your Relevant Equation.

What he said. The area of the shell is a function of $r$. Hence $A(r)$.