# Where is the Electric Field + Potential zero? Draw a graph

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1. Jun 22, 2013

### Reprisal35

1. The problem statement, all variables and given/known data
A -1.0 nC charge and a 1.0 nC charge are separated by 10.0 cm.
a) At what position(s) is the electric field due to the two charges zero?
b) At what position(s) is the electric potential due to the two charges zero?
c) Draw a graph showing the zero electric field and zero potential positions.

2. Relevant equations

$E = \frac{kq}{r^2}$

$V= \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

3. The attempt at a solution
I was able to figure out the answer for part a, the position in which the electric field is zero between the two charges, and got an answer of 5cm doing the following steps:

$$\frac{kq_1}{r_1^2} = \frac{kq_2}{r_2^2}\\ \frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}\\ \frac{q_1}{r_1^2} = \frac{q_2}{(10-r_1)^2}\\ \frac{1}{r_1^2} = \frac{1}{(10-r_1)^2}\\ r_1^2 = (10-r_1)^2\\ r_1^2 = r_1^2 - 20r_1 + 100\\ -20r_1 + 100 = 0\\ -20r_1 = -100\\ r_1 = 5$$

For part b, I got the same answer using a similar approach:

$$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}\\ V = \frac{q_1}{r_1} + \frac{q_2}{r_2}\\ V = \frac{q_1}{r_1} + \frac{q_2}{10-r_1}\\ 0 = \frac{-1}{r_1} + \frac{1}{10-r_1}\\ \frac{1}{r_1} = \frac{1}{10-r_1}\\ r_1 = 10-r_1\\ r_1 = 10 - r_1\\ 2r_1 = 10\\ r_1 = 5\\$$
Now for my problem. In part b, I'm not sure if I used the correct formula. I did some research on this and in a couple of the example, instead of changing $r_2$ to $10-r_1$, they just replaced $r_1$ with $10$ which ends up giving an answer of -10cm. The other examples did exactly what I did, including my class notes.

Assuming the method I used is correct, to answer part c, I would create a point, $Q_1$, and draw two circle around it at equal distance from each other. The inner circle will be 5cm from $Q_1$ and will be labeled as "zero electric field and zero potential". The outer circle will be 10cm from $Q_1$ and I would place another point along said circle and label it $Q_2$.

Example (Not photoshopped to scale):

I guess what I'm asking is, did I mess up somewhere because it seems too convenient that part a and b are the same and I'm not entirely sure if the graph is correct. Since it's asking for position(s), and 5cm is just the distance from a point, it should be in any direction from the point, and not locked into one specific location, thus the circle. Am I right to assume this?

2. Jun 22, 2013

### TSny

Hello Reprisal35 and welcome to PF!

Your answer to part (a) is incorrect. You need to consider the direction of the field from each charge. Remember, E is a vector quantity.

3. Jun 22, 2013

### Reprisal35

I forgot that they were opposite signs. What is now throwing me off is that since they are the same charge, but opposite sign, will there be two points, one in region I, and one in region III in which the electric field will be zero? Assuming this, the points where the electric field will be zero are -5cm, and +15cm? This feels wrong to me.

4. Jun 22, 2013

### TSny

If you're in region I and you want the two fields to cancel, then how would the magnitude of one field compare to the magnitude of the other field. If this is to happen, how would the distance r from one charge compare to the distance from the other charge? Is this possible at any point in region I? In region III?

Last edited: Jun 22, 2013
5. Jun 22, 2013

### Reprisal35

Wouldn't the magnitude of both the fields be the same but have opposite vectors in order to cancel two fields? This is possible in region I based on an example we did in class, but the example was of two charges with the value of 1nC and -2nC. I just can't seem to make the same connection with equal but opposite charges. It shouldn't be possible in region II since they have the same vector, my other choices are region I and region III. Based off of examples alone, the location where the electric field is zero is usually close to where the smaller magnitude is. Is it possible that the location of where the electric field is 0 is located at the point of both charges, aka 0cm and 10cm?

EDIT: I'll be back later. Need to take a break, been working on this since this morning. Thanks in advance for any advice though!

Last edited: Jun 22, 2013
6. Jun 22, 2013

### TSny

You just about have it. You need to keep thinking about whether or not it is possible for the fields from the two charges to have the same magnitude and opposite directions anywhere in region I or region III when the charges have the same magnitude.

Since the charges have the same magnitude, the only way the fields can have the same magnitude at a point is for the point to be the same distance from each charge.

7. Jun 22, 2013

### Reprisal35

Honestly, it just seems impossible for the fields from the two charges to have the same magnitude and opposite directions anywhere in region I or region III when the charges have the same magnitude because the magnitude isn't strong enough to go beyond the other charge. So either there's no point where the electric field is at 0, or it is at the mid point of the two charges, aka 5cm. This corresponds to your statement of "the only way the fields can have the same magnitude at a point is for the point to be the same distance from each charge", which also points back to my original answer of 5cm. So what am I misunderstanding here?

8. Jun 22, 2013

### TSny

Nothing. You've answered the question! The only places where the fields from the two charges have the same magnitude are the points on the midplane between the charges. But at none of those points do the fields point in opposite directions. So, there is no place where Enet = 0 (unless you want to count points at infinity).

9. Jun 22, 2013

### Reprisal35

Wow, it feels like the weight of my shoulder is gone now. Is there way I can express this in a formula, or it just based off of observation since the charges are equal but of opposite signs?

This kinda makes me wonder if the rest of the work I did for part b is also wrong. I want to say that part b is correct, but after going through part a and finally understanding why its nonexistent, I'm not so sure about electric potential. All the examples we covered in class and homework were based on two completely different charges, and not equal but opposite charges.

10. Jun 23, 2013

### TSny

For part (b) there are an infinite number of points where V is 0. You found one of them. Can you see where the others are? (No need to worry about vectors here, you just need V from each charge to have equal magnitudes and opposite signs. r2 does not have to equal 10 - r1.)

11. Jun 23, 2013

### Reprisal35

Normally, I would say they are 5cm around the original charge, but that's based off of the question giving us a set charge located a fixed distance from a given point and asking us where to place the second charge to create an electric potential of zero. This question gives us two charges and ask us to locate where the point will have an electric potential zero. Assuming the formula I used was correct to this extent:

$0 = \frac{q_1}{r_1} + \frac{q_2}{r_2}$

$\frac{-q_1}{r_1} = \frac{q_2}{r_2}$

$\frac{-(-1)}{r_1} = \frac{1}{r_2}$

$\frac{1}{r_1} = \frac{1}{r_2}$

$r_1 = r_2$

The only way this condition can be met is if $r_1$ and $r_2$ are of the same value and sign, meaning an infinite number of values. But since the restriction is that both charges are at 10cm from each other, I would assume that $r_1$ and $r_2$ are exactly 5cm from the point where the electric potential is 0. Since we aren't given a set location of where $q_1$ and $q_2$ are located, it can be any two distance on a plane as long as they are 10cm apart from each other. So if $q_1$ is at x=20, then $q_2$ is in any direction 10cm away from $q_1$ and the point where the electric potential is zero is exactly the midpoint between the two charges. Is this what you meant when you said it can be an infinite number of locations, right?

12. Jun 23, 2013

### TSny

You've shown that you need r1 = r2.

Suppose we introduce a coordinate system such that q1 is fixed on the x-axis at x = -5 cm and q2 is fixed on the x-axis at x = +5 cm. What would the total potential be at some point on the y-axis, say at y = +12 cm?

13. Jun 23, 2013

### Reprisal35

So the position where the electric potential is equal to zero is the vertical line at the point where x=0cm if q1=-5cm and q2=5cm?

14. Jun 23, 2013

### TSny

Yes, but there are a lot more points where V = 0. What about going out along the z-axis?

15. Jun 23, 2013

### Reprisal35

It should be the same as the y-axis if I'm picturing this correctly. It's as if there's a huge piece of paper that is stretching for infinity along the y and z-axis. That piece of paper represents the locations where the electric potential is zero.

16. Jun 23, 2013

### TSny

Yes, that's right. V = 0 everywhere on the yz plane.

17. Jun 23, 2013

### Reprisal35

Man, I can't believe I was wrong for both parts, thanks for showing me my mistakes and the correct method of solving this.

I can prove that part b is infinite along the y and z-axis by using a graph + a demonstration I just found on wolfram. In order to prove that part a is impossible, should I state that the answer I got, 5cm, is within region II which makes it impossible for the said charges to have a location where the electric field is at zero?

18. Jun 23, 2013

### TSny

There are a couple of ways you could argue this. You could say that the fields from the 2 charges must have the same magnitude in order to cancel each other. The only points where the fields have the same magnitude is where r1 = r2, which are the points on the yz plane. But for any of these points, it should be clear that the fields do not point in opposite directions. So, there are no points where the fields cancel.

Or, you could argue this way. The fields must point in opposite directions to cancel each other. The only points where the fields have opposite directions are on the x-axis in region I and region III. But you also need the fields to have the same magnitudes, and there are no points on the x-axis in I or III where the fields have the same magnitude because any of these points is closer to one of the charges than it is to the other charge.

19. Jun 23, 2013

### Reprisal35

I think I'll use a variation of what I said and your second argument, thanks again!

20. Jun 23, 2013

### TSny

You're welcome.