Question

really confused here....dielectric constant....

Answer 1

the lab manual says to plot capacitance vs delta x and find a linear fit. i did that. then it says k(dielectric constant) = b (from y=mx+b) *d/epsilon *A. my distance is 7 mm. my area is 252 mm^2. and logger pro says b is 25.6 mm. i plug that in and get a k value of 8.036*10^10....it's paper, so the k should be 3.5. what in the world happened?

Answer 2

I'm not really completely sure just by looking, but my first instinct would tell me to check your math.

You have all your measurements in mm, did you perhaps use epsilon which has meters rather than mm? I'm not sure that would make a huge difference, but it could possibly be something.

Answer 3

yeah, that could account for the huge difference in power, but not the value. it's still gonna be 8.036, not 3.5. that's still a big gap.

Answer 4

and it only drops it to 8.03*10^7 instead of 10^10

Answer 5

@Kainui, any ideas?

Answer 6

I mean, I don't really know what you're talking about to help you. Can you show your work or explain the experiment?

Answer 7

it's dealing with capacitors. this is what it says;

In the Dielectric “Material” window select paper from the drop down menu. As you slide the dielectric out of the capacitor, take readings of the capacitance C and the offset ∆x for 5 positions of the dielectric material in between the plates. In Logger Pro program make a plot of C (y-axis) vs. ∆x (x-axis). Apply linear fit to your set of data and from the intercept of this linear fit find the dielectric constant of the paper along with its error. How does your experimental value of the dielectric constant for paper compare with the known value of 3.5.
Attach the graph from Logger Pro to the lab report.

Answer 8

It is apparent that eqn. 6 is a linear function of ∆x (same as y=m∆x+b) with intercept
b = 〖 ε〗_0 A/d κ .
Then κ= bd/(〖 ε〗_0 A) , and if we know the uncertainty in the intercept ∆b from the Logger Pro fit, we can propagate it into the error in dielectric constant: ∆ κ=d/(〖 ε〗_0 A) ∆b.

Answer 9

@Kainui does it make sense now?

Answer 10

@ganeshie8 ? maybe? someone?

Answer 11

hello?????????????????????/

Answer 12

You originally said that you are plotting capacitance vs the length of the dielectric that is inserted between the plates of the capacitors. In in plot the magitude of the values you are using are not correct. Capacitance is measure in Farads and for a typical capacitor that should be of the order of a microfarad so your ordinate on your graph should be like \[Y \times 10^{-6}\] The consistent unit for distance is the meter so the abscissa should be in meters which means that the numbers should be of the order of \[X \times 10^{-3}\]

So the scale on your graph are incorrect which would give an incorrect slope.

Answer 13

well, i tried to do the capacitance of 1.1*10^-11, but logger pro doesn't see that as an actual number, so i just used .11. the scale shouldn't matter as long as the numbers are consistent. either way, i'm getting 8 instead of 3, which is still way too far off.

Answer 14

WOW I think the wrong is to get linear approximation for fractional function as
C= O/d .How does this give you a line of negative slope!!!
it would be
take the inverse of d to get linear function and use line.
C=O*(x) . x=1/d.
I believe if you took the inverse of dx the answer will be right :D

Answer 15

And the graph will be be increasing

Answer 16

@saiken2009
Are you here?

Answer 17

yeah. ran to start a load of laundry. here now

Answer 18

well, as the dielectric (paper) moves furhter away, the capacitance goes down, that's why slope is like that.

Answer 19

the manual says to plot C vs d. nothing else. not C vs 1/d.

In the Dielectric “Material” window select paper from the drop down menu. As you slide the dielectric out of the capacitor, take readings of the capacitance C and the offset ∆x for 5 positions of the dielectric material in between the plates. In Logger Pro program make a plot of C (y-axis) vs. ∆x (x-axis). Apply linear fit to your set of data and from the intercept of this linear fit find the dielectric constant of the paper along with its error. How does your experimental value of the dielectric constant for paper compare with the known value of 3.5.
Attach the graph from Logger Pro to the lab report.

Answer 20

But that can not be line with negative slope.
this is a fractional function.
Use the inverse of dx and get the answer then judge :D

Answer 21

Hey man logically : What is the value that make (..Any numbers..)/d = 0.

Answer 22

doing it that way, the b (intercept) is zero. which doesn't make any sense either

Answer 23

Hey we need just slope to calculate the k
slope = K*e_0*a .
use this to get K.

Answer 24

but i have to do it the way they tell me to do it in my lab report.

Answer 25

It is apparent that eqn. 6 is a linear function of ∆x (same as y=m∆x+b) with intercept
b = 〖 ε〗_0 A/d κ .
Then κ= bd/(〖 ε〗_0 A) , and if we know the uncertainty in the intercept ∆b from the Logger Pro fit, we can propagate it into the error in dielectric constant: ∆ κ=d/(〖 ε〗_0 A) ∆b.

Answer 26

This is math.I don't know what is the relation of k to any "intersect point" using program simulation which would get you the exact value as the law.

Answer 27

Let's us just use what I said earlier and get the answer to see if that is right.
If it is right just use this method and ask your instructor.

Answer 28

i don't even understand the purpose of this part of the lab.....in the previous part, we found k just by dividing the capacitance with the dielectric by the capacitance without....

Answer 29

well, the method you said earlier gives a K of zero cause the b is zero. 0/anything is 0.

Answer 30

I said use slope
k= slope/(e_0*a...)

Answer 31

what's a? the same as A? the area?

Answer 32

Yea

Answer 33

I am really sure this time that answer will be right :)

Answer 34

-107....

Answer 35

none of this makes any sense. why does it have to be so complicated to find K which is simply supposed to be a scalar.

Answer 36

hey First it can't be negative value !!!
secondly use SI units.

Answer 37

well, if the slope is negative, you're gonna have a negative answer. and i can't do the 1/d, because there's nothing in the lab manual about doing that or even using your formula. i have to follow the lab manual.

Answer 38

i know what K is. the point is trying to find it using their formula.

Answer 39

not to simply find K cause we already know it.

Answer 40

I am sure that you used wrong values.
(1/d) is directly with c.
You can do what ever you want I just want you to get the right answer, but you are making mistakes and telling me that is nonsense.

Answer 41

The aim of exper. is to find k.

Answer 42

the aim is to see how close that formula is to the actual value.

Answer 43

we already know k

Answer 44

And the manual is wrong to say that there would be y intercept 00= 1/d!!

Answer 45

the manual doesn't say that. i did it in excel. got an intercept of 1

Answer 46

er 0 using 1/d

Answer 47

which doesn't make sense to me to do it that way anyway because we are measuring capacitance over the change in distance. not the inverse of the change in distance.

Answer 48

Hey man you are using a program simulation not actual component to calculate error.
The programmer used the same formula as you did, but nature ?

Answer 49

what?

Answer 50

the dielectirc constant of paper is just a known value across the board. it's not programmed in the sim at all.

Answer 51

I mean your experiment.

Answer 52

you can plot the values which you enter and take fractional function approximation and see if it is the same as f(x)=1/x or not

Answer 53

are you provided with the relationship between C and delta?

Answer 54

charge =C*deltax

Answer 55

sorry, that's wrong.

Answer 56

i am now so much more confused than i was when i started....

Answer 57

That says if deltx =0 then C=0 which is not true.

Answer 58

that should have been charge = C*delta V. i have no idea about C and delta x

Answer 59

What are you suppose to plot?

Answer 60

By the way the capacitance for this set up with no dielectric is about 3.16E-10 Farads