# Finding E-field within a semiconductor.

• datran
In summary: Yes, you're doing fine here. You have found the E-field for 0<x<x0 and you have found that it is negative for any value of x in this region. That's good.The second part where -Xo < x < 0.I used the same equation as previous except exchange ρ1 with -ρ1 and used E(-Xo) = 0and got that E(x)= (ρ1/ epsilon)(x+Xo). This is 0 at the -Xo boundary, but again, E-field is pointing to the rightHowever, 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
datran

## Homework Statement

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

## Homework Equations

I used the ∇. E = ρ / epsilon

## 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:
Welcome back to PF.

datran said:

## Homework Statement

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

## Homework Equations

I used the ∇. E = ρ / epsilon

## 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?
Looks good so far. Note that E is an increasing function of x here, provided that ρ1 is positive.
If so, I got the whole E-field for 0 < x < Xo to be (ρ1/epsilon) [Xo - x]
Not quite. You get the correct E=0 at x=x0, but now E has a negative slope. This is easily fixed.

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

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:

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:
datran said:

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.
Yes, correct
When 0 < x < Xo,

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

∇·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.

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

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?
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.

The second part where -Xo < x < 0.

I used the same equation as previous except exchange ρ1 with -ρ1 and used E(-Xo) = 0
That's right. Except that you should use

∇·E = ρ/ε = -ρ1

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!
You're welcome.

Your approach is correct. To find the constant C, you can plug in the boundary condition at x = Xo, where E = 0. This will give you C = -Xo(ρ1/epsilon). Plugging this value of C into your equations, you get:

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

Your equations are correct. However, it is important to note that these equations only hold for the region -Xo < x < Xo. Outside this region, the electric field will be 0.

## What is the purpose of finding the E-field within a semiconductor?

The E-field, or electric field, within a semiconductor helps to determine the movement of charge carriers and the overall behavior of the material. This information is crucial for understanding how electronic devices made from semiconductors will function.

## How is the E-field within a semiconductor measured or calculated?

The E-field within a semiconductor can be measured or calculated using various techniques, such as Hall effect measurements, capacitance-voltage measurements, and numerical simulations using computer software.

## What factors affect the E-field within a semiconductor?

The E-field within a semiconductor is affected by several factors, including the type and concentration of dopants, the material's bandgap, and the applied voltage or current. Temperature, impurities, and defects can also influence the E-field.

## Why is it important to consider the E-field within a semiconductor in device design?

The E-field within a semiconductor has a significant impact on the performance and reliability of electronic devices. It affects carrier mobility, current flow, and potential barriers, which can affect device speed, power consumption, and overall functionality.

## How do variations in the E-field within a semiconductor impact device behavior?

Variations in the E-field within a semiconductor can lead to changes in device behavior, such as changes in electrical characteristics or even device failure. This is why accurately measuring and understanding the E-field is crucial in device design and production processes.

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