Finding electric potential at a point between 3 electrodes.

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Homework Statement:

The figure is an edge view of three charged metal electrodes. Let the left electrode be the zero point of the electric potential. Assume that Q=100 nC and q=50 nC.

The left electrode has a charge of -q, the middle electrode has a charge of +Q, and the right electrode has a charge of -q. There is 1 cm in between the left electrode, the middle electrode is 1 cm wide, and there is 1 cm between the middle electrode and the right electrode. All of the electrodes have 2.0 cm x 2.0 cm cross section. I apologize for the lack of a visual, I realize this would make it easier.

a.) What are E at 0.5 cm
b.) What are V at 0.5 cm
c.) What are E at 1.5 cm
d.) What are V at 1.5 cm
e.) What are E at 2.5 cm
f.) What are V at 2.5 cm

Relevant Equations:

E=Q/(2e_0*A)
V=Ed
30_P41.jpg

Firstly, I am not a English speaker. So I apologize that I cannot use English well..

I got a), c), e)
a)
at 0.5cm, E = -q/(2e_0*A) - Q/(2e_0*A) + q/(2e_0*A) = -1.4*10^7 V/m
c)
at 1.5 cm, E = 0 (inside electrode)
e)
at 2.5cm, E = -q/(2e_0*A) + Q/(2e_0*A) + q/(2e_0*A) = 1.4*10^7 V/m

And I am confused when calculating below..
I got b), f)
b)
at 0.5cm, V = E*d = 1.4*10^7 * 0.005 = 7*10^4 V
f)
at 2.5cm, V = E*d = 1.4*10^7 * 0.005 = 7*10^4 V

Are they right?

And, I don't know how to get the d) answer..
I think V = 0, because E = 0. Is it right?
 

Answers and Replies

  • #2
kuruman
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Why is the answer to (e) the negative of the answer to (a)? The expressions are the same. Also, if someone looks at the picture from the back of the screen, it would look exactly the same with left-to-right interchanged. Therefore fields and potentials at 0.5 cm and 2.5 should be identical by symmetry
 
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  • #3
TSny
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The equation V = Ed would be better written as ΔV = -Ed. The left side is the change in potential when moving a distance d in the direction of the electric field for a uniform field.

As a warm-up exercise, consider the following figure:
1572189324263.png


If the potential at ##b## is given to be 600 V, what are the potentials as ##a##, ##c## and ##d##.?
 
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  • #4
kuruman
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The equation V = Ed would be better written as ΔV = -Ed. The left side is the change in potential when moving a distance d in the direction of the electric field for a uniform field.
Better yet, ΔV = -E Δx where E is the magnitude of the field; this takes care of the cases when one moves opposite to the electric field.
 
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  • #5
rude man
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My approach to this kind of problem is hinted at as follows:

1. Assign surface charge densities s1, s2 .... s6 to the 6 surfaces from left to right. Assume unit area for all electrodes (the area is immaterial for purposes of introductory physics).
2. You can eliminate 2 of them right away by Gauss.
3. Then, three equations are just saying that surface charges for each electrode sum to the respective element's charge.
4. The 4th equation is determined by summing forces on a test charge inside one electrode (where the E field is zero) due to all six surface charges and equating to zero. This is perhaps the non-obvious part.
5. Now one can solve for all 6 surface charges, giving of course the six D fields, E fields, and by integration, potentials.

Although this particular problem is easily solved by symmetry considerations, in general all three (or more) electrodes can have differing charges so the above method offers a general solution.
 
  • #6
rude man
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Why is the answer to (e) the negative of the answer to (a)?
The OP is correct. The E fields point in opposite direction about the middle electrode. The charge on both surfaces is the same, viz. Q/2, but the E fields point oppositely.
 
  • #7
kuruman
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The OP is correct. The E fields point in opposite direction about the middle electrode. The charge on both surfaces is the same, viz. Q/2, but the E fields point oppositely.
Yes, the OP is correct. My objection was that the algebraic expressions leading to the numerical answers in (a) and (e) do not differ by an overall negative sign while the numerical answers do. In my post I said the expressions are the same because I missed reading the minus sign in front of Q in the first equation.
 
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