Resistors and equipotential lines?

[edgarpokemon]

Homework Statement

We are suppose to make a lab report about and experemint that we did in physics lab. It consisted of drawing equipotential lines in a overbeck electric field mapping apparatus. Different desing plates were used, each having conductors, some were circular and other where just straight figures. I am not worried too much about the design of the equipotential lines or how they are arranged, I am more concerned about the resistors that were on the apparatus. We seted up the power source to 4 volts, and we had to find points of 0 voltage on the map by the use of a galvanometer, it is called a potential difference. so my question is this, how does resistors affect equipotential lines?

Homework Equations

The Attempt at a Solution

https://www.google.com/search?q=overbeck+field+mapping+apparatus&rlz=1C1VFKB_enUS623US623&source=lnms&tbm=isch&sa=X&ved=0ahUKEwi44orP-ofPAhVLMyYKHck7CxgQ_AUICCgB&biw=1366&bih=640#imgrc=8XMPWlgsynBtdM:

that is a diagram of the apparatus that we used. In lectures we have not talked even about equipotential lines and resistor so that is why i am very confused. I read the book and I understand the concepts of both but I am not sure what is going on on the apparatus. We first connected the galvanometer to resistor E1 of the apparatus to find a 0 reading, my assumption is that since it is only a resistor, the voltage will not be decreased as much, so the equipotential line will be close to the charge particle. As we move to E2, the voltage should have decreased since there are now 2 resistors, and so on. As the equipotential lines get farther and farther away from the charged particle, the potential (voltage) decreases. Is this right?

 

[Merlin3189]

I'm not sure what your problem is here.
What do you think the resistors are for?
How did YOU map the equipotential lines?

looking at the instructions in your link, I see,

With moderate pressure between the probe contact and the black surface of the plate, move the probe to various points between A and B until a point is found where G reads zero and mark this point through the hole in the top of the probe . Finding this null point may be tedious because G is very sensitive and slight variations such as probe pressure may cause it to fluctuate. If it fluctuates between + and - (zero is surely somewhere in between!), this probably is an adequate "zero" reading. Practice finding the null points until a satisfactory technique has been achieved. Realize that each null point found is at the same potential (voltage) as is E4 because no current flows from E4 to the point (P). Thus, all null points found while G is connected to E4 are at the same potential and, therefore, form an equipotential line.

The last two sentences say it all! So perhaps you can expand on your question?

 

[edgarpokemon]

I'm not sure what your problem is here.
What do you think the resistors are for?
How did YOU map the equipotential lines?

looking at the instructions in your link, I see,
The last two sentences say it all! So perhaps you can expand on your question?

I'm not sure what your problem is here.
What do you think the resistors are for?
How did YOU map the equipotential lines?

looking at the instructions in your link, I see,
The last two sentences say it all! So perhaps you can expand on your question?

So all resistors will cause the reading of the voltage difference to be 0? There were seven resistors labeled E1 to E7. And resistor 1 caused the equipotential lines to be closer to the charge. Resistor 2 caused the equipotential lines to get farther away from the charge, and so on. Resistors are to reduce the voltage current and prevent a short circuit. So will resistor E1 reduce all the 4 volts that we applied to 0? so resistors prevent current from flowing. I have no idea about electrical circuits. I have seen circuits with 1 resistor but not 7 and i am pretty confused!?

 

[edgarpokemon]

I'm not sure what your problem is here.
What do you think the resistors are for?
How did YOU map the equipotential lines?

looking at the instructions in your link, I see,
The last two sentences say it all! So perhaps you can expand on your question?

http://alpcentauri.info/equipotential_lines.html

we got equipotential lines similar to those on the right side (with the curved electric fields). each equipotential line has a different resistor, so imagine labeling each equipotential line, the one closer to the charge as E1, and so on. so what is the function of the resistors?

 

[edgarpokemon]

I'm not sure what your problem is here.
What do you think the resistors are for?
How did YOU map the equipotential lines?

looking at the instructions in your link, I see,
The last two sentences say it all! So perhaps you can expand on your question?

http://tinypic.com/r/2cfbli1/9

here is one that we did. my question is, why are equipotential lines separated by a distance x , from resistor to resistor? like why is E1 closer than E2 to the circle?

 

[Merlin3189]

Ok. So the resistors are there to provide different potentials. Assuming the 8 resistors in this diagram are all the same, then the potentials the chain provides are; 1/8 of 4V, 2/8 of 4V, 3/8 of 4V, ... 7/8 of 4V. 0V and 4V are the two electrodes themselves.
So if we said A was at 0V, then B would be at 4V and the resistor junctions would be at; E1=0.5V, E2=1V, E3=1.5V,...,E7= 3.5V. The resistors form a potential divider.

So when the galvanometer is connected to say E3, then all the ploted points would have a potential difference from A of 1.5V (and a PD from B of 2.5V) and would all be at the same potential (ie. equipotential.)

This only works for points where the galvanometer reads zero, because then we know that no current is flowing from the resistor chain through the galvanometer. The potential divider is based on the idea that the same current flows through each resistor in the chain. If the current is the same and the resistance is the same, then the voltage (PD) across each resistor is the same - 1/8 , 2/8, 3/8, ... of the full voltage difference across the ends. So long as no current enters or leaves part way along, then the current is the same in all of them. The actual amount of resistance is not significant, so long as they're all the same and no current flows out in the middle.

The other side of the galvanometer story is that, if there is no current flowing through the galvanometer, then the potential must be the same at both ends. So P is at whatever potential the resistor junction is.

here is one that we did. my question is, why are equipotential lines separated by a distance x , from resistor to resistor? like why is E1 closer than E2 to the circle?

Looks good to me.
E1 is at 1/8 of 4V = 0.5V potential difference from A. So all the points you plot when connected to E1 are also at 0.5V difference from A.
Then you connect to E2 at 2/8 of 4V = 1V pd from A. So all the points you plot now are also at 1V pd from A.
A is at the lowest potential and the pd from A increases as you get nearer to B. And B is at the highest pd from A, 4V.

Where your electrodes are points, then the PDs are in the same ratio as the distances from the electrodes. The line down the middle is equidistant from A and B, and it has a 2V pd from A and a 2V pd from B. The line where you are 1V PD from A and 3V PD from B, then you are 3x as far from B as from A. The ring round the electrode is not a circle, because when you are on the opposite side of A from B, then you are further from B than on the near side. So you also have to be further from A.

 

[edgarpokemon]

Ok. So the resistors are there to provide different potentials. Assuming the 8 resistors in this diagram are all the same, then the potentials the chain provides are; 1/8 of 4V, 2/8 of 4V, 3/8 of 4V, ... 7/8 of 4V. 0V and 4V are the two electrodes themselves.
So if we said A was at 0V, then B would be at 4V and the resistor junctions would be at; E1=0.5V, E2=1V, E3=1.5V,...,E7= 3.5V. The resistors form a potential divider.

So when the galvanometer is connected to say E3, then all the ploted points would have a potential difference from A of 1.5V (and a PD from B of 2.5V) and would all be at the same potential (ie. equipotential.)

This only works for points where the galvanometer reads zero, because then we know that no current is flowing from the resistor chain through the galvanometer. The potential divider is based on the idea that the same current flows through each resistor in the chain. If the current is the same and the resistance is the same, then the voltage (PD) across each resistor is the same - 1/8 , 2/8, 3/8, ... of the full voltage difference across the ends. So long as no current enters or leaves part way along, then the current is the same in all of them. The actual amount of resistance is not significant, so long as they're all the same and no current flows out in the middle.

The other side of the galvanometer story is that, if there is no current flowing through the galvanometer, then the potential must be the same at both ends. So P is at whatever potential the resistor junction is.Looks good to me.
E1 is at 1/8 of 4V = 0.5V potential difference from A. So all the points you plot when connected to E1 are also at 0.5V difference from A.
Then you connect to E2 at 2/8 of 4V = 1V pd from A. So all the points you plot now are also at 1V pd from A.
A is at the lowest potential and the pd from A increases as you get nearer to B. And B is at the highest pd from A, 4V.

Where your electrodes are points, then the PDs are in the same ratio as the distances from the electrodes. The line down the middle is equidistant from A and B, and it has a 2V pd from A and a 2V pd from B. The line where you are 1V PD from A and 3V PD from B, then you are 3x as far from B as from A. The ring round the electrode is not a circle, because when you are on the opposite side of A from B, then you are further from B than on the near side. So you also have to be further from A.

aaaaa I understand now! in lecture we have not talked about this and they already want us to know! thank youuu :P