Question

!!Urgent Question!! Help would be greatly appreciated.

What is the capacitance of a 50um diameter cell with a 7nm thick cell wall whose dielectric constant is 9.0? Because the cell's diameter is much larger than the wall thickness, it is reasonable to ignor the curvature of the cell and think of it as a parallel-plate capacitor.

Answer 1

@Fifciol not sure what i'm doing wrong... would you mind helping?

Answer 2

Yes, you can consider this capacitor as parallel-plate with circular area.
\[C=\epsilon_r \epsilon_{0}\frac{ A }{l }=\epsilon_r \epsilon_{0}\frac{ \pi d^2 }{4l }\]

Answer 3

where does that equation come from? and what does epsilon sub r equal?

Answer 4

From the definition of capacitance C=q/V and Gauss's law. Epsilon sub r is dielectric constant.

Answer 5

ok Can you show the derivation, I'm kind of lost.

Answer 6

What does L equal?

Answer 7

definition of capacitance:
C=q/V
Gauss's law:
\[\epsilon_0 \epsilon_r ES=q\]El=V
Substitute and you'll get the result

Answer 8

l is thickness

Answer 9

If I just substitue, where does A come from?

Answer 10

A is S in gauss's law sorry for that

Answer 11

Sorry, but could you re-write the equation? I'm lost with all the variables...

Answer 12

\[\epsilon_0\epsilon_rEA=q\]

Answer 13

Ok, got. Thanks, I appreciate your patience.

Answer 14

yw

Answer 15

How would I solve for area??

Answer 16

You assume that area is a circle so its pi r^2 = pi d^2/4

Answer 17

word diameter speaks for itself

Answer 18

l=thickness
A=area
E=Electric field
q= charge
V=potential
d=diameter

Answer 19

that's correct V is potential difference

Answer 20

Very nice! @Fifciol

Answer 21

thanks @abb0t :)

Answer 22

I got 2.1e-11 F, but that is the wrong answer.

Answer 23

\[C=\frac{ (\kappa A \epsilon _{0}) }{ d }\] where K=dielectric constant, A=area Epsilon=8.85e-12, and d=cell thickness

Answer 24

I got 22.33 * 10^-12

Answer 25

0.022 nF

Answer 26

How?

Answer 27

is that your book answer?

Answer 28

\[C=\frac{ (9)( \pi )(2.5e-5)^2 }{ (7e-9) }\]

Answer 29

no book answer...

Answer 30

you forgot the epsilon zero
8.85 e-12

Answer 31

.022 nF is wrong also.

Answer 32

what is book answer?

Answer 33

I used epsilon zero in my equation, just didnt write it.

Answer 34

Book doesn't give answer.

Answer 35

Then how do you know it is wrong?

Answer 36

online HW

Answer 37

would surface area have anything to do with it?

Answer 38

@theEric

Answer 39

I don't know I can help with this one, sorry. But I will tell you that the area would be important because that is where the charges on opposite poles will gather, and the electric field there has something to do with the potential, I think.

Is this like a capacitor whose films wind around? Like, they have the one contact, then dielectric, then other contacts sandwiched together and then rolled up like a sleeping bag?

Then you might be able to find area somehow.

@Fifciol , what do you think the shape of the capacitor might be? I think you found the area of a circle, so do you think that there are two circular plates? I haven't had a problem like this before.

Answer 40

Spherical?

Answer 41

Yes I consider it as two parallel circular plate

Answer 42

THen what would be the equation? It should be what was worked out above. But that answer is incorrect

Answer 43

maybe try 22.34 pF

Answer 44

@Fifciol , that makes it easier! Also suggestion. Sometimes online questions are pick about their answers... I haven't checked the math yet, though.

Answer 45

tried that.

Answer 46

Dang! Maybe the units of \(\epsilon_0\) are an issue? I have to go, sorry. Good luck!

Answer 47

http://session.masteringphysics.com/myct/itemView?assignmentProblemID=27760944