Capacitance nd Spherical cells

In summary, the conversation discusses the concept of parallel conducting plates with charges +Q and -Q separated by a distance b. It addresses questions about the charge per unit area on each plate, the electric field between the plates, the capacitance, and the effect of inserting a dielectric between the plates. The second question involves the diffusion of oxygen in air and water and asks for a comparison of the diffusion distance L. The conversation also includes attempts at solving the problems and seeking clarification.
  • #1
SallyBieber
13
0
Hey all I'd like to help me..
thnx in advance
here z tha Q's :

1) Two plane parallel conducting plates each have area S and are separated by a distance b.
One carries a charge +Q ; the other carries a charge -Q. Neglect edge effects

a) What is the charge per unit area on each plate ? where does it reside ?
b) what is the electric field between the plates ?
c) what is tha capacitance ?
d) As the plate separation is increased, what happens to E, v and C ?
e) If a dielectric is inserted b/w the plates, what happenes to E, v and C ?

.....
2) The distance L that oxygen can diffuse in the steady state is approximately L= √(CD/Q)
, where C is the oxygen concentration, D is the diffusion constant, and Q is the rate per unit volume that oxygen is used for metabolism

a) show that L has dimensions of length
b) The diffusion of oxygen in air is about 10,000 times larger than the diffusion of oxygen in water . By how much will the diffusion distance L change if oxygen diffuses through air instead of water, all other things being equal ?
 
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  • #2
You should try to solve the problem yourself first. Sorry, but physics forums has rules and I might get banned if I do the problem for you. For the first question, are you required to derive the result using Gaussian surfaces, or simply to use the equation for a capacitor? Either way, try to solve the question, and post your attempt.
 
  • #3
well okay I'll try .. but please I need ur correction
the 1st one bout simply eq. for capacitor
..
so az I learned the first one its bout
( charge density ( sigma ) = charge (q) / area ( S ) ) and cus its bout one plate I think I've to dividin the equ. by 2 !
so it 'll be like
Sigma = q/2*S
reside ?? humm well in the middle between the 2 plates :\

b)Electric field will be
E= Sigma/ e_0 ; while sigma = Q/S
so E= Q/S*e_0
c) C= Q/V = Q/E*B = Q*S*e_0/Q*b = S*e_0/b
but I've got another relation which is C = kappa * e_0 * S / b
which one should I take ?

d)
C will be decreased according to C = e *S/b
and the E stay the same I think! and the potential will increased

e) dielectric will increased tha C , and decreased the E .. and the potential I don't know :\
 
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  • #4
please guys any hint for the 2nd Q ??
 
  • #5
I'v doubts bout tha 2nd Q ..
C = m-3
D = m2/s
Q= m-3/s
so L = m ..
is that right !
while D grows by 10,000
" assume the remaining is constant "
L = 100
is that tha better way to explain tha part B in Q 2 ?
 
  • #6
SallyBieber said:
so az I learned the first one its bout
( charge density ( sigma ) = charge (q) / area ( S ) ) and cus its bout one plate I think I've to dividin the equ. by 2 !
so it 'll be like
Sigma = q/2*S
reside ?? humm well in the middle between the 2 plates :\

b)Electric field will be
E= Sigma/ e_0 ; while sigma = Q/S
so E= Q/S*e_0
c) C= Q/V = Q/E*B = Q*S*e_0/Q*b = S*e_0/b
but I've got another relation which is C = kappa * e_0 * S / b
which one should I take ?

d)
C will be decreased according to C = e *S/b
and the E stay the same I think! and the potential will increased

e) dielectric will increased tha C , and decreased the E .. and the potential I don't know :\

a) The question says that the absolute value of the total charge per plate is Q, so you don't need to divide by 2. The question said the two plates were conducting, so for a conductor, where does the extra charge go? (I'm hoping you've been taught this).

b)Yep, this is right.

c)The first one is correct. I think the second one is using kappa as the relative permittivity, which would be used if there was a material between the plates.

d)Yep, all correct.

e)They are correct, but do you know why? Think of what the dielectric material does.
 
  • #7
a) the charge will be in the inside of each plates
so I'm goin to take the first relation which is Sigma = Q/S
am I right ?

e) the dielectric materials prevent tha charges to flow through the materials so the Q will stay the same and doesn't make any change on the E
right ?
 
  • #8
SallyBieber said:
I'v doubts bout tha 2nd Q ..
C = m-3
D = m2/s
Q= m-3/s
so L = m ..
is that right !
while D grows by 10,000
" assume the remaining is constant "
L = 100
is that tha better way to explain tha part B in Q 2 ?

You've done part a correctly. Part b is worded badly. It says "The diffusion of oxygen in air is about 10,000 times larger" And you've taken that to mean D is 10,000 times larger. I think you're probably right, but the question is vague.
 
  • #9
e) and the potential v will be decreased right ?
 
  • #10
BruceW said:
You've done part a correctly. Part b is worded badly. It says "The diffusion of oxygen in air is about 10,000 times larger" And you've taken that to mean D is 10,000 times larger. I think you're probably right, but the question is vague.

so u think I've done part B too ? any suggest ?
 
  • #11
SallyBieber said:
a) the charge will be in the inside of each plates
so I'm goin to take the first relation which is Sigma = Q/S
am I right ?

e) the dielectric materials prevent tha charges to flow through the materials so the Q will stay the same and doesn't make any change on the E
right ?

a) Sigma = Q/S is correct. And yes, the charge is confined to the plates, but think where the charges will want to go (it won't be in the middle of the plate).

e)No, you got it right before when you said E will be less. Dielectric materials contain bound charges. you're right that there is no flow of charge, but the bound charges in the dielectric material have the ability to align in a certain way. There is an electric field going through the dielectric, so how do you think the bound charges will align themselves? And what effect does this have on the total electric field?
 
  • #12
SallyBieber said:
so u think I've done part B too ? any suggest ?

The question is worded badly, since 'diffusion is 10,000 times larger' is a very vague statement, and it could mean anything. But in the context of the question, it is most likely talking about D, so I think you did it right.
 
  • #13
BruceW said:
a) Sigma = Q/S is correct. And yes, the charge is confined to the plates, but think where the charges will want to go (it won't be in the middle of the plate).

e)No, you got it right before when you said E will be less. Dielectric materials contain bound charges. you're right that there is no flow of charge, but the bound charges in the dielectric material have the ability to align in a certain way. There is an electric field going through the dielectric, so how do you think the bound charges will align themselves? And what effect does this have on the total electric field?

maybe the charges will be distribute in each plates
or maybe it will be going toward the +ve charge ..

when the dielectric material placed the E its causing Dielectric polarization Because of dielectric polarization, positive charges are displaced toward the field and negative charges shift in the opposite direction

but the potential when the dielectric material placed will be decreased because of wht ?
 
  • #14
SallyBieber said:
maybe the charges will be distribute in each plates
or maybe it will be going toward the +ve charge ..

when the dielectric material placed the E its causing Dielectric polarization Because of dielectric polarization, positive charges are displaced toward the field and negative charges shift in the opposite direction

but the potential when the dielectric material placed will be decreased because of wht ?

a)I'm not sure what you mean by 'it will go toward the +ve charge'... There are +ve charges on one plate and -ve charges on the other plate, so just use the rule that opposites attract, and where do the charges end up?

e)Yes, the positive charges are displaced in the direction of the field and the negative charges in the opposite direction. So what effect does this have on the total electric field? And to find the potential, just use V=Eb
 
  • #15
BruceW said:
a)I'm not sure what you mean by 'it will go toward the +ve charge'... There are +ve charges on one plate and -ve charges on the other plate, so just use the rule that opposites attract, and where do the charges end up?

e)Yes, the positive charges are displaced in the direction of the field and the negative charges in the opposite direction. So what effect does this have on the total electric field? And to find the potential, just use V=Eb

will be goes toward the plate with -ve Q ..

e) the total will be zero
 
  • #16
I still don't understand 'goes toward the plate with -ve Q', sorry.
e)Well, the bound charges will not be enough to totally cancel out the electric field, so the electric field will be less, but not zero.
 
  • #17
if the charge will reside inside each plates .. that's mean the E will be moving from the +ve charge plate to the -ve charge plate
so there is no specific place for the charge !
is it like this way ?
 
  • #18
The charge is inside each plate yes. And yes, the E field goes from the +ve plate to the -ve plate. I'm not sure what you mean by no specific place for the charge..

To try to explain it: let's say plate A has some extra positive charges and plate B has some extra negative charges. The charges in plate A will be attracted to the charges in plate B, so the charges in plate A move as close as they can to plate B, so they end up on the surface of plate A which is closest to plate B. And a similar thing will happen for the charges in plate B.
 
  • #19
well yeah I get it
thank you thank you so much
really appreciate it and thank you for your patient ..
Really thank you
my problems are solved ..
thank you again
 
  • #20
Cool. All's well that ends well.
 

1. What is capacitance?

Capacitance is a measure of an object's ability to store an electrical charge. It is defined as the ratio of the charge stored on an object to the potential difference (voltage) across the object.

2. How is capacitance calculated?

Capacitance is calculated using the formula C = Q/V, where C is the capacitance in Farads (F), Q is the charge in Coulombs (C), and V is the potential difference in Volts (V).

3. What is a spherical cell?

A spherical cell is a type of capacitor that consists of two concentric spheres, with the inner sphere acting as the positive plate and the outer sphere acting as the negative plate. It is commonly used in high-voltage applications due to its compact design and high capacitance.

4. How does the size of a spherical cell affect its capacitance?

The capacitance of a spherical cell is directly proportional to the surface area of its plates. Therefore, a larger spherical cell will have a higher capacitance compared to a smaller one, as it has a larger surface area for charge storage.

5. What factors affect the capacitance of a spherical cell?

The capacitance of a spherical cell is affected by the distance between the two plates, the dielectric material between the plates, and the surface area of the plates. It is also influenced by the type of material used for the plates, as well as the voltage applied to the cell.

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