Find the drift speed of of the electrons

[r-swald]

Homework Statement

A solid, conductive, rectangular block with a resistance of 150 Ohm's has a cross-sectional area of .11m across it's width and a length of .2m. The block contains 1.2x10^19 conductive electrons, and a connected battery creates a 25v potential.

Relevant Equations

I = qVnA
I = V / R
n = # conductive electrons / m^3

I've found I to be .167 using the potential and resistance.

I also found the volume by multiplying the cross-sectional area by the length (?) and then dividing the # of conducting electrons into that to find packing density (n).

To find drift speed, I would also need the area of the block as well as the charge. I'm not sure wether the block is equal in it's height and width to use .11/2 to find those to use in the area equation.

 

[etotheipi]

Consider a cross-section of the block at some point along its length. We'll assume that the charge carriers "drift" as one homogenous medium of charge with $n = 1.2 \times 10^{19}$ electrons per cubic metre, at a drift velocity of $v_d$.

In a time $\delta t$, the charge will have advanced $v_d \delta t$. So if we multiply this by the cross sectional area, we obtain the total volume of this charge-filled medium that has passed through our cross section, namely $V = Av_d \delta t$. It follows that the total charge that has passed through our cross section is $Q = neAv_d \delta t$, since $ne$ is just the charge per unit volume.

But remember, current is defined as the rate of flow of charge past a given point in the circuit (in this case, through our cross sectional surface!). So $I = \frac{neAv_d \delta t}{\delta t} = neAv_d$. This was the equation you gave in your relevant formulae section, and is in fact all you need to solve the question from what you have already worked out.

Can you rearrange that equation for $v_d$?

 

[r-swald]

Consider a cross-section of the block at some point along its length. We'll assume that the charge carriers "drift" as one homogenous medium of charge with $n = 1.2 \times 10^{19}$ electrons per cubic metre, at a drift velocity of $v_d$.

In a time $\delta t$, the charge will have advanced $v_d \delta t$. So if we multiply this by the cross sectional area, we obtain the total volume of this charge-filled medium that has passed through our cross section, namely $V = Av_d \delta t$. It follows that the total charge that has passed through our cross section is $Q = neAv_d \delta t$, since $ne$ is just the charge per unit volume.

But remember, current is defined as the rate of flow of charge past a given point in the circuit (in this case, through our cross sectional surface!). So $I = \frac{neAv_d}{\delta t} = neAv_d$. This was the equation you gave in your relevant formulae section, and is in fact all you need to solve the question from what you have already worked out.

Can you rearrange that equation for $v_d$?

Ok, so Vd = I / nA
is there a difference between n and ne?

I'm still not sure how to find the area to plug into this equation

 

[etotheipi]

Ok, so Vd = I / nA
is there a difference between n and ne?

Yes. A factor of $e$. It should be $v_d = \frac{I}{neA}$.

I'm still not sure how to find the area to plug into this equation

Which area makes the most sense? I.e. which one did I use in the derivation ?

 

[r-swald]

Yes. A factor of $e$. It should be $v_d = \frac{I}{neA}$.

Which area makes the most sense? I.e. which one did I use in the derivation ?

I don't get it. Ok so I'm trying to find the drift speed of the electrons, I use this by dividing current (.167) by ne • A. You used cross-sectional area in your derivation so I'd use that, but I'm not sure what the factor of e is so I tried the packing density I found by dividing the "n" you used by volume (.022)

Vd = .167 / 5.45e20 • .11
= 2.79x10^-21

Not sure what I'm missing here, my professor didn't explain this one well

 

[etotheipi]

You used cross-sectional area in your derivation so I'd use that,

This is correct. As for the other part,

$n$ is the number density of electrons, the number of electrons in one cubic metre. This varies in different materials. $e$ is the elementary charge, $1.6\times 10^{-10} C$.

Now $n$ has units of per cubic metre, and $e$ has units of charge. $ne$ has units of charge per cubic metre, $C m^{-3}$! It follows that $ne$ times the volume is the total charge.

It's just a constant.

 

[hutchphd]

There is a dimensional error in your statement of the problem...please identify it. And can we require units on the calculations please...

 

[r-swald]

This is correct. As for the other part,

$n$ is the number density of electrons, the number of electrons in one cubic metre. This varies in different materials. $e$ is the elementary charge, $1.6\times 10^{-10} C$.

Now $n$ has units of per cubic metre, and $e$ has units of charge. $ne$ has units of charge per cubic metre, $C m^{-3}$! It follows that $ne$ times the volume is the total charge.

It's just a constant.

1.6x10^-19? I used that number and it worked. I was also using the wrong number for n, it was given as the total number of charged particles within the block and then divided by the volume was the n I was looking for. I'd also converted cm to m incorrectly, but it all worked out in the end :) thanks for the help! those equations were super helpful