Estimates of voltage drop with distance in weak electric field

In summary: If you put the dipole moment along the ##z##-axis, where is the point where you measured the voltage? On the axis, in the ##xy## plane or somewhere in between?
  • #1
livio
11
1
Homework Statement
I have an underwater dipole (interpole distance is small, probably 1 cm, but cannot be estimated easily so this can only be an approximated measure) generating an electric field which voltage can be measured at 0 cm and is equal to roughly 5 mV. The reading is referred to a reference electrode placed far away at a distance of 100 cm. I would like to estimate the voltage readings at increasing distances for the values 0, 1, 2, 4, 8, 16, 32, 64,…256 cm. According to a simple decay law |E|∼W(t)/d^3 the magnitude of the electric field read at distance d should decay with the distance cube. So based on this, I have estimed these values

voltage drop (2nd column) over distance (1st column)
2 1
4 0,125
8 0,015625
16 0,001953125
32 0,000244141
64 3,05176E-05
128 3,8147E-06
256 4,76837E-07

with the first column being the distance in cm and the second the voltage in mV. However the starting point is kind of arbitrary and it starts from a reading of 2 mV. I would like to start from 5 mV and make sure I am calculating the decay properly (when does the first drop start and how? if for instance I set the first distance at 1 cm is it fair to expect a 8-fold drop already there?). I guess one way would be to resolve the field equation of the dipole at different distances (0 and 1 cm) but I wonder which equation would be correct to use. Would the following be correct

E(R) = [3(p⋅R^)R−p] / [4πε*R^3]

E should be the electric field at the point individuated by the vector R. p is the dipole moment. R is the distance (or the length of the vector).
Relevant Equations
E(R) = [3(p⋅R^)R−p] / [4πε*R^3]
If I resolve the equation in 0, imposing a voltage value of 5 mV, it gives a non real solution, therefore I cannot resolve it for R=1 because I do not know which voltage value to impose. I am sure this is simpler than I am putting it :) thanks for any advice!
 
Physics news on Phys.org
  • #2
Welcome to PF.

Do you have a sketch or picture of the test setup? Why would the potential and E-field of a small dipole be referenced to another electrode somewhere else?

Edit -- You can upload a diagram using the "Attach files" link below the Edit window.
 
  • #3
Thank you Berkeman :) the reference far away is because the reading of 5 mV was taken in this way (not by me) so I am just reporting. I guess what is relevant is that the measuring electrode is not in "differential mode" but is referred to some ground far away, enough not to be affected by the dipole. I do not have a sketch right now but... imagine just a dipole "object" of 15-20 cm length within which the poles (positive and negative) may not be that far away... maybe 1 cm, maybe less ... but this actually should not be critical. What I know is that I start from 5 mV and I would like to estimate how the voltage drops. For simplicity one could estimate a dipole size of either 15 cm or 1 cm and just calculate the drop in both cases. My problem is that using the equation I have... if I bring R to 0 I definitely go nowhere :) at least... not as far as I would like to
 
  • #4
What are you trying to determine with your measurements? I assume it is the dipole moment. Since you have voltage measurements, why bother with the electric field? Use the voltage. The potential of a dipole relative to infinity is given by $$V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$$where ##p## is the dipole moment and ##\theta## is the angle between the direction of the position vector of the measurement and the direction of the dipole, i.e. ##\cos\theta=\dfrac{\mathbf{p} \cdot\mathbf{r}}{pr}## which I assume you know because the direction matters. From what you have said, your reference electrode is far enough to be considered infinitely far away.
 
  • #5
kuruman said:
What are you trying to determine with your measurements? I assume it is the dipole moment. Since you have voltage measurements, why bother with the electric field? Use the voltage. The potential of a dipole relative to infinity is given by $$V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$$where ##p## is the dipole moment and ##\theta## is the angle between the direction of the position vector of the measurement and the direction of the dipole, i.e. ##\cos\theta=\dfrac{\mathbf{p} \cdot\mathbf{r}}{pr}## which I assume you know because the direction matters. From what you have said, your reference electrode is far enough to be considered infinitely far away.

thank you. so just to make sure I understood: using the voltage the decay is with distance square, using the field the decay is with the cube ? is the field the first derivative of the voltage over space? regardless of this, I basically resolve it for r = 0 and obtain p, then use the same value to resolve it for the other distances...right?
 
  • #6
livio said:
I basically resolve it for r = 0 and obtain p, then use the same value to resolve it for the other distances...right?
I am not sure what you mean by this. There are two quantities you should be concerned with, the distance from the dipole ##r## and the angle ##\theta## with respect to the direction of the dipole. If you put the dipole moment along the ##z##-axis, where is the point where you measured the voltage? On the axis, in the ##xy## plane or somewhere in between? The angle with respect to the ##z##-axis of the point of measurement matters if you want to make sense of the results. The value of the potential on a hypothetical sphere with the dipole at the center depends on the latitude. Relative to infinity, it is positive at the north pole (##\theta =0##), zero at the equator (##\theta =\frac{\pi}{2}##) and negative at the south pole (##\theta=\pi##). At an arbitrary latitude, it has in between values.

More correctly, the field is the gradient (three-dimensional derivative) of the potential, $$\mathbf{E}=-\mathbf{\nabla}V=-\left(\frac{\partial V}{\partial x}\mathbf{\hat x}+\frac{\partial V}{\partial y}\mathbf{\hat y}+\frac{\partial V}{\partial z}\mathbf{\hat z}\right).$$ That is why direction matters.
 
  • Like
Likes berkeman
  • #7
kuruman said:
I am not sure what you mean by this. There are two quantities you should be concerned with, the distance from the dipole ##r## and the angle ##\theta## with respect to the direction of the dipole. If you put the dipole moment along the ##z##-axis, where is the point where you measured the voltage? On the axis, in the ##xy## plane or somewhere in between? The angle with respect to the ##z##-axis of the point of measurement matters if you want to make sense of the results. The value of the potential on a hypothetical sphere with the dipole at the center depends on the latitude. Relative to infinity, it is positive at the north pole (##\theta =0##), zero at the equator (##\theta =\frac{\pi}{2}##) and negative at the south pole (##\theta=\pi##). At an arbitrary latitude, it has in between values.

More correctly, the field is the gradient (three-dimensional derivative) of the potential, $$\mathbf{E}=-\mathbf{\nabla}V=-\left(\frac{\partial V}{\partial x}\mathbf{\hat x}+\frac{\partial V}{\partial y}\mathbf{\hat y}+\frac{\partial V}{\partial z}\mathbf{\hat z}\right).$$ That is why direction matters.
my problem is to calculate the voltage drop at different distances ( even along 1 direction only) given a known value at 0 cm (so touching the dipole)
 
  • Skeptical
Likes berkeman
  • #8
Looks to me that you have estimated the voltages according to an inverse cube law. If so, you are confusing potential and field strength.

Since the interpole distance is relatively large, don't use the usual dipole approximation. Just derive it from first principles. At distance x>d/2 from the dipole's centre, what is the potential due to each pole? What is the net potential?
 
  • #9
haruspex said:
Looks to me that you have estimated the voltages according to an inverse cube law. If so, you are confusing potential and field strength.

Since the interpole distance is relatively large, don't use the usual dipole approximation. Just derive it from first principles. At distance x>d/2 from the dipole's centre, what is the potential due to each pole? What is the net potential?
you are right, thank you for noting that. I can't measure at x>d/2. I only have an average measurement "along" the dipole and this is 5mV. If with net potential you mean the voltage measured across the dipole and not referenced to the a distant reference electrode, this should be around 1-2 mV.
The inter-pole distance is overestimated when I say 1 cm. It could be much less. So in general I know that the cube law is accepted in such cases (this is why I have used it). I just need to refer it to field (mV/cm) and not voltage (mV). what I should do is probably resolve the field equation in 0 and 1 cm using the voltage reading at 0 (V = 5 mV) to obtain a voltage reading at 1 cm within that field. Correct? :) thank you for your replies anyway, they helped me clear some doubts for sure.
 
  • #10
livio said:
I can't measure at x>d/2
I wasn't suggesting to measure it. I meant if the dipole charges are +q and -q, distance d apart, what should the potential be at x > d/2 from the middle (along the line of the dipole)?
livio said:
I only have an average measurement "along" the dipole and this is 5mV. If with net potential you mean the voltage measured across the dipole and not referenced to the a distant reference electrode, this should be around 1-2 mV.
Now I am confused. One would expect the potential to have a peak positive value at one pole and a negative value of equal magnitude at the other (relative to the potential at infinity). If the average along the dipole exceeds the potential difference across the dipole then there is some other field interfering.
 
  • #11
haruspex said:
I wasn't suggesting to measure it. I meant if the dipole charges are +q and -q, distance d apart, what should the potential be at x > d/2 from the middle (along the line of the dipole)?

Now I am confused. One would expect the potential to have a peak positive value at one pole and a negative value of equal magnitude at the other (relative to the potential at infinity). If the average along the dipole exceeds the potential difference across the dipole then there is some other field interfering.
what I know is that close contact readings on the dipole surface are between 2 and 20 mV with an average of 5 mV. Ignore the statement of the measurement "across" the dipole, that was an estimate of mine from another model. Sorry for the confusion. The problem is that whatever equation I am using, I can't resolve it in 0 otherwise the dipole moment goes also to 0 (unless I am doing the math wrong).
 
  • #12
livio said:
close contact readings on the dipole surface are between 2 and 20 mV with an average of 5 mV.
If the peak voltage differential you can find at the surface of the dipole container is 20mV then surely that indicates the axis of the dipole. The average differential from a set of random pairs of measurements would seem to be irrelevant.
livio said:
whatever equation I am using, I can't resolve it in 0 otherwise the dipole moment goes also to 0 (unless I am doing the math wrong).
Please post your working.
In the absence of a clear description or diagram, I am having to make guesses about the set-up.
My suggestion is that you should model the dipole as two point charges of unknown charge and unknown locations within the container. Write the expression for what the potential should be further out along the axis. How many actual measurements would you need to determine the unknowns?
 
  • #13
I hope this helps. I have modeled the dipole as you suggested using two point sources placed 15 cm apart. This would be "container", in reality the poles are probably much closer inside. I have placed the reference ground at 2 meters. I have a reading at distance 0 and need the other ones. The measurement does not have to be necessarily at the positive pole, they could be placed on the midline perpendicular to the dipole axis.
 

Attachments

  • sketch.png
    sketch.png
    5.8 KB · Views: 48
  • #14
I have no idea what the 5mV is measuring. Unless there is some other charge around (is that a battery you have marked at the bottom?), the voltage should be zero all along the midline. It will not be zero along the measurement line you have marked.
 
  • #15
the measurement should not be on the midline necessarily either. let's say the reading is variable around the dipole (the absolute voltage reading) but on average is around 5 mV. I would need to take that 5 and calculate the drop for increased distances. let's consider 5 mV the reading at an X position between + and -. Now I need to draw a perpendicular line passing by X and crossing the dipole axis and read voltage along the way.
 
  • #16
It might be helpful if you described exactly what you are measuring.. I know I am confused, and believe you are also. What does "I have an underwater dipole" mean???
 
  • #17
hutchphd said:
It might be helpful if you described exactly what you are measuring.. I know I am confused, and believe you are also. What does "I have an underwater dipole" mean???
it means there is a dipole, underwater, it can be modeled as two point charges separated 15 cm apart. an average of 5 mV voltage can be measured in close proximity to it. The voltage will drop with distance. I need to estimate the drop. see the sketch in the previous comments.
 
  • #18
livio said:
it can be modeled as two point charges separated 15 cm apart.
How do you know this?? Physically what are you modelling???
 
  • #19
hutchphd said:
How do you know this?? Physically what are you modelling???
since I do not know the actual pole distance but only the size of the container of the dipole (15 cm) I followed the suggestion to model it using a 2 point charge dipole (and I think this is accurate).
 
  • Sad
Likes hutchphd
  • #20
Hi @livio. Some miscellaneous points….

Is the dipole producing a constant or an oscillating electric field? From what’s been said, let’s assume it’s constant.

The potential along the ‘midline’ of a dipole is zero as shown here for example:
https://i.stack.imgur.com/0hQ6Q.png
(diagram from: https://physics.stackexchange.com/questions/339425/potential-nonsense).

Water consists of polar molecules and mobile ions (assuming you are using liquid water). Placing an electric dipole in water will cause positive charge to build up around the negative end of the dipole and vice versa. You are essentially embedding the dipole in a conducting medium, not in a simple dielectric. The usual 'simple dielectric' equations won't apply.

By the way, your description (Post #1) says that the reference point for voltage measurements is 100cm from the dipole. But your Post #13 diagram shows it much further than this.
 
  • Like
Likes berkeman
  • #21
Steve4Physics said:
Hi @livio. Some miscellaneous points….

Is the dipole producing a constant or an oscillating electric field? From what’s been said, let’s assume it’s constant.

The potential along the ‘midline’ of a dipole is zero as shown here for example:
https://i.stack.imgur.com/0hQ6Q.png
(diagram from: https://physics.stackexchange.com/questions/339425/potential-nonsense).

Water consists of polar molecules and mobile ions (assuming you are using liquid water). Placing an electric dipole in water will cause positive charge to build up around the negative end of the dipole and vice versa. You are essentially embedding the dipole in a conducting medium, not in a simple dielectric. The usual 'simple dielectric' equations won't apply.

By the way, your description (Post #1) says that the reference point for voltage measurements is 100cm from the dipole. But your Post #13 diagram shows it much further than this.
thanks :) this is getting far too complicated than I thought.... I hope not. I think what I need is a simple equation to estimate the drop in voltage. this obviously depends on water resistivity too (this can be estimated around 5 kOhm/cm). thank you for noting the incongruency of the distance, the thing is, because the medium is water, that distance is close to "infinity" and I will not reach it with my measurements (or am I wrong on this?). I hope this helps. In any case, I do not want to stress anyone :) I will try to figure it out.
 
  • #22
livio said:
..what I need is a simple equation to estimate the drop in voltage
But the potential is zero everwhere along the dipole's midline - see post #20. Taking V=0 at infinity, the potential does not change as you move along the midline; it remains zero. So there should be no voltage-drops. Or are you looking at points which are not on the midline?

livio said:
this obviously depends on water resistivity too
That's only true if a current flows through the water. For static charges (which are what you apparently have) there is no current - so the resistivity of water is irrelevent here.

I think the underlying issue is that you haven't told us enough about the system or the purpose of the whole experiment. For example, what actually is the dipole - some sort of electret? Do you know the charge? Is this a real experiment requiring a large tank of water or just hypothetical?
 
  • Like
Likes hutchphd

Related to Estimates of voltage drop with distance in weak electric field

What is voltage drop?

Voltage drop is the decrease in voltage that occurs when electrical current flows through a conductor. It is caused by the resistance of the conductor and can be calculated using Ohm's law.

What is a weak electric field?

A weak electric field is an electric field with a low intensity. It is typically measured in volts per meter (V/m) and is weaker than the electric field produced by high voltage power lines or household appliances.

How does distance affect voltage drop in a weak electric field?

The voltage drop in a weak electric field is directly proportional to the distance traveled by the electrical current. This means that the farther the current travels, the greater the voltage drop will be.

What factors can affect voltage drop in a weak electric field?

Aside from distance, the resistance of the conductor, the strength of the electric field, and the amount of current flowing through the conductor can also affect the voltage drop in a weak electric field.

Why is estimating voltage drop important in scientific research?

Estimating voltage drop is important in scientific research because it allows scientists to understand and predict the behavior of electrical currents in different materials and environments. This information can be used to design and optimize electrical systems, as well as to ensure the safety and efficiency of electrical devices.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
335
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Electrical Engineering
Replies
14
Views
533
  • Introductory Physics Homework Help
Replies
1
Views
722
Replies
19
Views
1K
Replies
38
Views
3K
Back
Top