Question Regarding Electric Fields

  • #1
Greetings everyone,

I have a question about determining electric fields when given parallel planes of charge. In particular, I need some help getting the sign conventions straight in my head. I've attached 2 quick problems with solutions in a Word document so that perhaps you can help me to spot my errors. Thanks in advance!!
 

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  • #2
Doc Al
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For question 1: Your diagrams look good, but you got the wrong answer. The equation:
[tex]E = \frac{\eta}{2 \epsilon_0}[/tex]
gives you the magnitude of the field (treat eta as just the magnitude of the surface charge); use the sign of the charge to find the direction of the field. You have the direction of the fields shown perfectly in your diagram. In region 1, two of the field contributions are negative (down) and one is positive (up). They cancel.

If you want to have eta in that equation include the sign of the charge, then you must treat it as a positive charge when drawing the vectors. (If eta is negative, it will switch the direction for you.)

When you write: E1 = -Ep + Ep' -Ep'', you are implying that Ep, Ep', & Ep'' are just the magnitudes of the fields. You already attached the needed signs in front of those numbers. Yet you also put in "-eta" when calculating the fields, which canceled the signs!

In question 2, since you are solving for the surfaces charges, you took into account the fact that eta2 was negative, so you used "-eta2" when calculating the field. Perfectly OK.

Hope that helps a little.
 
  • #3
Still Confused

Hi Doc Al,

Thank you very much for your reply. I greatly appreciate the help. I give you a lot of credit for reading through my long post! :redface:

I read through your explanation and I now understand the first question (question 46 in the attachment below), but it still doesn't seem that my instructor followed the same sign conventions throughout his solutions and I don't quite understand why he made an exception in the equation for region 2 in question 48 (I've given more details below). I've attached both the questions (there are 2, question 46 and question 48) along with the instructor's solutions.

Some thoughts: In both questions 46 and 48, the diagrams are broken up into regions 1 - 4 around, or inside of, the planes of charge. The direction of the vectors [tex]\vec{E}_p[/tex] , [tex]\vec{E}_{p^'}[/tex] , and [tex]\vec{E}_{p^''}[/tex] can be determined by considering the charge (+ or -) on each of the planes. If the plane is positively charged, then its electric field vector [tex]\vec{E}[/tex] in each of the regions points away from this plane. If the plane is negatively charged, then its electric field vector [tex]\vec{E}[/tex] in each of the regions points toward this plane.

When writing the equations for each of the regions 1 - 4 in the solutions to both questions 46 and 48, the sign associated with the magnitude of each of the vectors [tex]\vec{E}_p[/tex] , [tex]\vec{E}_{p^'}[/tex] , and [tex]\vec{E}_{p^''}[/tex] is determined by the direction of their respective vectors in the diagram - if [tex]\vec{E}[/tex] is pointing up, [tex]\frac{\eta}{2\epsilon}[/tex] [tex]\hat{j}[/tex] is positive; if [tex]\vec{E}[/tex] is pointing down, [tex]\frac{\eta}{2\epsilon}[/tex] [tex]\hat{j}[/tex] is negative.

My instructor has followed this same sign convention for every equation except for one, the equation for region 2 in question 48. This is the only region that falls within a plane, in this case a conductor. However, since the surface charges on the conductor are polarized (positive on the top surface and negative on the bottom surface), it seems that I can just treat question 48 as though it is 3 planes of charge (just as in question 46) - with a negative plane of charge between 2 positive planes of charge.

Now, if I follow the same sign conventions in the equation for region 2 (question 48) as were followed in every other equation, then I should get

[tex]\vec{E}_p+\vec{E}_{p^'}[/tex] + [tex]\vec{E}_{p^''}[/tex] = [tex]\frac{-\eta_1}{2\epsilon}\hat{j}-\frac{\eta_2}{2\epsilon}\hat{j}+\frac{\eta_3}{2\epsilon}\hat{j}=0[/tex]

So [tex]-\eta_1-\eta_2+\eta_3=0[/tex]


The equation I got above cannot be correct because this, along with the equation [tex]\eta_1+\eta_2=0[/tex] for the neutral conductor, would lead the erroneous result of [tex]\eta_3=0[/tex], which obviously cannot be true.

Furthermore, this is not the equation obtained by my instructor in his solution (see attached). Is this because region 2 is within a conductor? Does it have something to do with the statement "Let [tex]\eta_1[/tex] , [tex]\eta_2[/tex] , and [tex]\eta_3[/tex] be the surface charge densities of the three surfaces with [tex]\eta_2[/tex] a negative number" in the solution? I am truly perplexed. Any help would be GREATLY appreciated! Thank you!
 

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  • #4
Doc Al
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Some thoughts: In both questions 46 and 48, the diagrams are broken up into regions 1 - 4 around, or inside of, the planes of charge. The direction of the vectors [tex]\vec{E}_p[/tex] , [tex]\vec{E}_{p^'}[/tex] , and [tex]\vec{E}_{p^''}[/tex] can be determined by considering the charge (+ or -) on each of the planes. If the plane is positively charged, then its electric field vector [tex]\vec{E}[/tex] in each of the regions points away from this plane. If the plane is negatively charged, then its electric field vector [tex]\vec{E}[/tex] in each of the regions points toward this plane.
Good.

When writing the equations for each of the regions 1 - 4 in the solutions to both questions 46 and 48, the sign associated with the magnitude of each of the vectors [tex]\vec{E}_p[/tex] , [tex]\vec{E}_{p^'}[/tex] , and [tex]\vec{E}_{p^''}[/tex] is determined by the direction of their respective vectors in the diagram
I agree with this.
- if [tex]\vec{E}[/tex] is pointing up, [tex]\frac{\eta}{2\epsilon}[/tex] [tex]\hat{j}[/tex] is positive; if [tex]\vec{E}[/tex] is pointing down, [tex]\frac{\eta}{2\epsilon}[/tex] [tex]\hat{j}[/tex] is negative.
But not this. If E points up (and you're using the usual convention that up is positive) then the field will be positive. But [itex]\eta/2\epsilon[/itex] can be positive or negative, depending on the sign of the charge. And [itex]\eta/2\epsilon[/itex] is just a number--it doesn't know if you are above or below anything.

For example, let's say the charge density on a given surface is positive (thus [itex]\eta > 0[/itex]). Above the surface the field is [itex]\eta/2\epsilon[/itex], but below the surface the field is [itex]-\eta/2\epsilon[/itex]. You have to put in the correct sign.

My instructor has followed this same sign convention for every equation except for one, the equation for region 2 in question 48. This is the only region that falls within a plane, in this case a conductor. However, since the surface charges on the conductor are polarized (positive on the top surface and negative on the bottom surface), it seems that I can just treat question 48 as though it is 3 planes of charge (just as in question 46) - with a negative plane of charge between 2 positive planes of charge.
Sure.

Now, if I follow the same sign conventions in the equation for region 2 (question 48) as were followed in every other equation, then I should get

[tex]\vec{E}_p+\vec{E}_{p^'}[/tex] + [tex]\vec{E}_{p^''}[/tex] = [tex]\frac{-\eta_1}{2\epsilon}\hat{j}-\frac{\eta_2}{2\epsilon}\hat{j}+\frac{\eta_3}{2\epsilon}\hat{j}=0[/tex]

So [tex]-\eta_1-\eta_2+\eta_3=0[/tex]
You are mixing things up. The charge densities are [itex]\eta_1, \eta_2, \eta_3[/itex]. Note that [itex]\eta_2 < 0[/itex], a negative number. Thus your field equation should be:

[tex]\vec{E}_p+\vec{E}_{p^'}[/tex] + [tex]\vec{E}_{p^''}[/tex] = [tex]-\frac{\eta_1}{2\epsilon}\hat{j}+\frac{\eta_2}{2\epsilon}\hat{j}+\frac{\eta_3}{2\epsilon}\hat{j}=0[/tex]

And your charge equation should be:

[tex]\eta_1+\eta_2=0[/tex]



The equation I got above cannot be correct because this, along with the equation [tex]\eta_1+\eta_2=0[/tex] for the neutral conductor, would lead the erroneous result of [tex]\eta_3=0[/tex], which obviously cannot be true.
Right.

Furthermore, this is not the equation obtained by my instructor in his solution (see attached). Is this because region 2 is within a conductor?
No.
Does it have something to do with the statement "Let [tex]\eta_1[/tex] , [tex]\eta_2[/tex] , and [tex]\eta_3[/tex] be the surface charge densities of the three surfaces with [tex]\eta_2[/tex] a negative number" in the solution?
That's the one that's messing you up. :wink:

Let me know if I have addressed the right issue.
 
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