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Finding E-field within a semiconductor.

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data
    rho(x) = 0 for x >= Xo and x <= -Xo
    rho(x) = ρ1 for 0< x < Xo
    rho(x) = -ρ1 for -Xo < x < 0.

    The last two rho's are constants.

    Electric field = 0 for x> Xo and x < -Xo.
    Find E for -Xo< x < Xo

    2. Relevant equations

    I used the ∇. E = ρ / epsilon

    3. The attempt at a solution

    Since it is ρ1 when x> 0 and -ρ1 when x< 0, I split it into two equations.

    I get x hat partial d/dx dotted with E -x hat = ρ1/ epsilon.
    So I got E(x) = x(ρ1/epsilon) + C. Do I plug in the rightmost boundary of Xo where E = 0 to find C? If so, I got the whole E-field for 0 < x < Xo to be (ρ1/epsilon) [Xo - x]

    Is my second equation correct:

    using the same ideas, x hat partial d/dx dotted with E -x hat = -ρ1 /epsilon. I use the condition that E(-Xo) = 0 and I get the equation of

    E-field for -Xo < x < 0: E = (ρ1/ epsilon)[x- Xo].

    E-field for 0< x < Xo: E = (rho1 / epsilon)[Xo-x]
    Thanks for any responses!
     
    Last edited: Jan 21, 2013
  2. jcsd
  3. Jan 21, 2013 #2

    Redbelly98

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    Welcome back to PF. :smile:

    Looks good so far. Note that E is an increasing function of x here, provided that ρ1 is positive.
    Not quite. You get the correct E=0 at x=x0, but now E has a negative slope. This is easily fixed.

    Slight problem: when x= -x0 -- that's negative x0 -- your expression is not zero.

    EDIT ADDED:
    Another thing to check in your final expressions: you have a negatively charged region to the left of a positively charged region. Therefore, which way should E point within the charged regions? Does this match up with your expressions? (I.e., your expressions are either positive if you think it's a rightward-pointing field or they are negative if it's a leftward-pointing field.)
     
    Last edited: Jan 21, 2013
  4. Jan 21, 2013 #3
    Thanks for replying!

    Since it is ρ1 when x> 0 and -ρ1 when x < 0 the E-field is pointing to the left.
    In my standard coordinate system, that means the E-field should be negative.

    When 0 < x < Xo,

    is it correct to state that (x hat partial d/dx)(-x hat E) ? is equal to ρ1 / epsilon?

    I am confused whether I have the correct vector E with the right unit vector.

    In doing so, I still arrive to the same conclusion (by plugging in that E(Xo) = 0 to be
    E(x) = (ρ1/epsilon)(Xo-x).

    So you're saying since the E-field is pointing to the left, therefore this quantity should be negative, but it isn't right? I don't know what I'm missing then.

    The second part where -Xo < x < 0.

    I used the same equation as previous except exchange ρ1 with -ρ1 and used E(-Xo) = 0
    and got that E(x)= (ρ1/ epsilon)(x+Xo). This is 0 at the -Xo boundary, but again, E-field is pointing to the right
    However, through inspection because of the negatively charged to the left and positively charged to the right, it should be pointing to the left (negative) rather than to the right (positive).

    Once again, thanks for the help!

    EDIT: If I don't incorporate the direction of x hat in the del . E = ρ /epsilon,

    I get E(x) = (ρ1/epsilon) (x-xo) where 0<x<xo
    and E(x) = (-ρ1/epsilon)(x+xo) where -xo<x<0.

    The second part of the question says to find the potential of V(xo) given that V(-xo) = 0.

    I used the equaiton V(xo) - V(-xo) = - integral E . dl

    and split it into two integrals from -xo to 0. But in this E, I incorporated E as integral(E -x hat . x hat dx)
    and added it to the second part using the same negative x hat. Am I using the unit vector correctly in these parts?

    Thanks!
     
    Last edited: Jan 21, 2013
  5. Jan 21, 2013 #4

    Redbelly98

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    Yes, correct :smile:
    Not quite. Why are you putting a minus sign in here? You had it right in your first post,

    ∇·E = ρ/ε

    so

    ∂E/∂x = ρ1

    There is no minus sign here. The negative comes into play after you solve this equation and have a final expression to look at. Then you can check whether that expression is negative.

    You have the correct unit vector x_hat. It is a one-dimensional problem, depending on x and not on y or z. x_hat is the only unit vector that is needed here.

    Correct, that is what I'm saying. You got a positive value for E, that's how you can tell that something went amiss.

    You were on the right track earlier, when you got

    E(x) = x(ρ1/epsilon) + C

    But the algebra got messed up when you tried to find C, so you just need to redo that part.

    That's right. Except that you should use

    ∇·E = ρ/ε = -ρ1

    You're welcome.
     
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