# Finding electric potential at a point between 3 electrodes.

• jangchen
In summary, the speaker apologizes for not being a native English speaker and provides a summary of their calculations for the given scenario. They are able to calculate the electric fields and potentials at different distances from the electrodes. They are unsure about one calculation and ask for confirmation. The speaker also addresses a question about the symmetry of the scenario and explains their approach to solving these types of problems. Finally, they clarify their reasoning for why the electric fields at 0.5cm and 2.5cm should be opposite in direction.
jangchen
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

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?

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

robot6
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:

If the potential at ##b## is given to be 600 V, what are the potentials as ##a##, ##c## and ##d##.?

scottdave
TSny said:
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.

scottdave and TSny
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.

kuruman said:
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.

rude man said:
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.

## 1. What is electric potential?

Electric potential is a measure of the electric potential energy per unit charge at a specific point in an electric field. It is also known as voltage.

## 2. How is electric potential calculated?

The electric potential at a point is calculated by dividing the electric potential energy at that point by the amount of charge present at that point.

## 3. What is the formula for finding electric potential at a point between 3 electrodes?

The formula for finding electric potential at a point between 3 electrodes is V = (V1Q1 + V2Q2 + V3Q3) / (Q1 + Q2 + Q3), where V represents the electric potential at the point and Q represents the amount of charge on each electrode.

## 4. How does the distance between electrodes affect the electric potential at a point between them?

The electric potential at a point between electrodes is inversely proportional to the distance between the electrodes. This means that as the distance between electrodes increases, the electric potential at a point between them decreases.

## 5. What is the significance of finding electric potential at a point between 3 electrodes?

Finding electric potential at a point between 3 electrodes is important in understanding the behavior of electric fields and the distribution of electric potential in a system. It can also be used to determine the direction and magnitude of electric current in a circuit.

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