The resistivity of a semiconductor can be modified by adding...

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length $L$ and cross-sectional area A lies along the $x$-axis between $x =$ 0 and $x = L$. The material obeys Ohm's law, and its resistivity varies along the rod according to $\rho(x) = \rho{_0}$ exp($-x/L$). The end of the rod at $x =$ 0 is at a potential $V_0$ greater than the end at $x = L$. (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude $E(x)$ in the rod as a function of $x$. (c) Find the electric potential $V(x)$ in the rod as a function of $x$. (d) Graph the functions $\rho(x)$, $E(x)$, and $V(x)$ for values of $x$ between $x =$ 0 and $x = L$.

Transcript

00:01 All right, so in this problem, so we have a rod.

00:07 And we know the cross section of the rod is a, and total length is l.

00:12 And the resistivity is a function of x, where x is defined from the left, from the left side to some position inside this rod.

00:21 So this is x, okay.

00:23 And row equal row nots, e to the negative l over l.

00:28 Okay.

00:34 So in part a, we want to find out the resistance.

00:36 Of this rod and also find out the current through this rod.

00:41 So we suppose the potential on the left side is v0s and the potential on the right hand side is 0.

00:47 So r, the resistance, sorry.

00:51 So the first step, we just look at the small piece of the rod whose lens is dx.

00:58 Then the resistance of this small piece is the r has the expression r times dx over a, right? and row has the expression right here.

01:10 So this is equal row not, e to the power negative l over l, negative x over l times dx over a, right? so the total resistance is equal integral dr, or we just plug in this expression, r0 not to the power negative l over l, dx over a, and x ranges from zero to l, right? and this expression gives you l ronaut over a minus e to the negative 1.

01:49 And the current inside this road equal v0 over r, right? so this is just the om's law.

01:58 So we have this a times v0 divided by l times row not and the 1 times 1 divided by e minus e to negative.

02:12 One, okay? and in part b, so we want to find out the the electrofield inside this road.

02:26 So basically, again, we just look at a small piece of the rod.

02:31 So we have dv equal i times d r, right? so dr has expression, sorry, dr and i.

02:43 They, we have found expression in this page.

02:47 So this is expression for i.

02:50 And this is expression for the dr, right? so we can plug in these two expressions.

02:55 And we see that this equal a v0 divided by l times the row not, and 1 divided by 1 minus negative 1 minus e to the negative 1 minus 1 ,000, and times row not times e to the power negative x over l, and times d x over a.

03:22 So, after.

03:25 After simplification, you see that this equal v0 over l times 1, divided by 1 minus e to the negative 1, e to the power negative x over l, and times dx.

03:44 So the definition of the electric field is vv over dx, all right? so this actually gives you the coefficient right here.

03:54 V0 over l, 1 minus, 1 minus e to the negative 1, e to the negative 1, e to the negative 1, e to the negative l of l.

04:04 So this is the electric field inside this road...

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