Find the capacitance of a half-full capacitor

In summary: Don't get fooled by the polarization field of the dielectric. The two electric fields ##\vec{E_d}## and ##\vec{E_u}## are equal, it is the respective displacement fields ##\vec{D_d}=\epsilon\vec{E_d}## and ##\vec{D_u}=\epsilon_0\vec{E_u}## that are...
  • #1
madafo3435
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Homework Statement
An isolated spherical capacitor has charge + Q in its interior (radius ra) and charge −Q in its outer conductor (radius rb). Then half the volume between the two conductors is filled with a dielectric liquid with constant K. I will attach a picture with the drawing.

Find the capacitance of the half-full capacitor.
Relevant Equations
The relevant equations are exposed in the procedure below.
The capacitance must be found, that is:

##C=\frac {Q}{V_a-V_b}##.​

Let ##\vec{E_d}## and ##\vec{E_v}## be the electric fields due to the zone with dielectric and the zone without dielectric respectively.

In the case of a spherical capacitor with a vacuum between its plates, it is easy to prove that:

##C_0=4\pi \epsilon _0(\frac {r_ar_b}{r_b-r_a})##
Now, here I don't know what to do, but I think the following: If the upper and lower spherical capacitors are at the same potential difference ##V_{ab}##. Then:

##V_a-V_b=\int_{r_a}^{r_b} \vec{E_d} \textbf{.} d\vec{r} = \int_{r_a}^{r_b} \vec{E_v} \textbf{.} d\vec{r} ##
Then the capacitor would be equivalent to two capacitors in parallel, ##C_d## y ##C_v## respectively. Each with half the charge compared to that of the spherical capacitor. Is obtained:

##C_d=\frac {kC_0}{2} = 2\pi K\epsilon _0(\frac {r_ar_b}{r_b-r_a})##

##C_v=\frac {C_0}{2} = 2\pi \epsilon _0(\frac {r_ar_b}{r_b-r_a})##
Therefore:

##C=C_v+C_d=2\pi \epsilon _0(1+K)(\frac {r_ar_b}{r_b-r_a})##
I don't know if this procedure is correct, and if it is, I have trouble proving that the potential difference due to each field mentioned above is the same.

I appreciate the help you can give.
 

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  • #2
I think you did all the hard work and you were drown in a spoon of water:wink:.
Isn't the potential of the inner spherical conductor everywhere on its surface the same? (and equal to ##V_a##)
What about the potential of the outer spherical conductor?
 
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  • #3
madafo3435 said:
I don't know if this procedure is correct, and if it is, I have trouble proving that the potential difference due to each field mentioned above is the same.
A conductor under static conditions is always an equipotential. The half of the inner shell that contains the dielectric is at the same potential as the half that contains the vacuum. Likewise for the outer shell. Therefore, the potential difference between any point on the inner shell and any other point on the outer shell is the same.

This problem is somewhat misleading in that the capacitance is a geometric property and does not depend on whether the capacitor is charged or not. It is irrelevant that it is isolated with some charge already on it when the dielectric is introduced. That becomes relevant if someone asked you to find the charge on each hemisphere.
 
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  • #4
Delta2 said:
I think you did all the hard work and you were drown in a spoon of water:wink:.
Isn't the potential of the inner spherical conductor everywhere the same? (and equal to ##V_a##)
What about the potential of the outer spherical conductor?
radially yes, but the field that the dielectric can generate confuses me :confused:
 
  • #5
kuruman said:
A conductor under static conditions is always an equipotential. The half of the inner shell that contains the dielectric is at the same potential as the half that contains the vacuum. Likewise for the outer shell. Therefore, the potential difference between any point on the inner shell and any other point on the outer shell is the same.

This problem is somewhat misleading in that the capacitance is a geometric property and does not depend on whether the capacitor is charged or not. It is irrelevant that it is isolated with some charge already on it when the dielectric is introduced. That becomes relevant if someone asked you to find the charge on each hemisphere.
I'm fine with what you say, but I'm still confused about something. I understand that the entire conductor is equipotential, but this is due to all the field generated by the two spheres and the dielectric. But if we want to find the power due to only one of the hemispheres, we would be considering a different field in each hemisphere, since one hemisphere has dielectric and the other does not. I don't know if I was clear with my concern ...
 
  • #6
madafo3435 said:
But if we want to find the power##\dots##
What power? What is the expression for it? Did you mean something else and got lost in translation?
 
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  • #7
kuruman said:
What power? What is the expression for it? Did you mean something else and got lost in translation?
yes, I'm sorry. I wanted to say: if we want to find the potential difference due to only one of the hemispheres, we see that the field is different in each hemisphere, because one hemisphere has dielectric and the other does not. Then, when calculating the potential difference due to different fields, the potential difference would be different. This is what I was thinking.
 
  • #8
madafo3435 said:
radially yes, but the field that the dielectric can generate confuses me :confused:
Don't get fooled by the polarization field of the dielectric. The two electric fields ##\vec{E_d}## and ##\vec{E_u}## are equal, it is the respective displacement fields ##\vec{D_d}=\epsilon\vec{E_d}## and ##\vec{D_u}=\epsilon_0\vec{E_u}## that are different.
 
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  • #9
madafo3435 said:
##\dots~##we see that the field is different in each hemisphere,##~\dots##
The fields may be different but the potential difference will be the same because charge ##Q## will be redistributed to make it so. Let ##C_0## = the capacitance without the dielectric. With the dielectric, as you have stated correctly, ##C_d=\kappa C_0/2## and ##C_v= C_0/2##. Let ##Q_d## and ##Q_v## be the total charge on each hemisphere. Note that ##Q_d+Q_v=Q.##

The condition for the potential difference to be the same is $$\frac{Q_d}{C_d}=\frac{Q_v}{C_v}~\Rightarrow~\frac{Q_d}{\kappa}=\frac{Q_v}{1}~\Rightarrow~Q_d=\kappa Q_v.$$ With the requirement that the sum of the charges be ##Q##, you get$$Q_v=\frac{1}{\kappa+1}Q~~\text{and}~~Q_d=\frac{\kappa}{\kappa+1}Q.$$ See how it works?

On Edit: This another way of saying what @Delta2 posted while I was composing this.
 
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  • #10
Delta2 said:
Don't get fooled by the polarization field of the dielectric. The two electric fields ##\vec{E_d}## and ##\vec{E_u}## are equal, it is the respective displacement fields ##\vec{D_d}=\epsilon\vec{E_d}## and ##\vec{D_u}=\epsilon_0\vec{E_u}## that are different.
interesting observation, his contribution enlightened me
 
  • #11
kuruman said:
The fields may be different but the potential difference will be the same because charge ##Q## will be redistributed to make it so. Let ##C_0## = the capacitance without the dielectric. With the dielectric, as you have stated correctly, ##C_d=\kappa C_0/2## and ##C_v= C_0/2##. Let ##Q_d## and ##Q_v## be the total charge on each hemisphere. Note that ##Q_d+Q_v=Q.##

The condition for the potential difference to be the same is $$\frac{Q_d}{C_d}=\frac{Q_v}{C_v}~\Rightarrow~\frac{Q_d}{\kappa}=\frac{Q_v}{1}~\Rightarrow~Q_d=\kappa Q_v.$$ With the requirement that the sum of the charges be ##Q##, you get$$Q_v=\frac{1}{\kappa+1}Q~~\text{and}~~Q_d=\frac{\kappa}{\kappa+1}Q.$$ See how it works?

On Edit: This another way of saying what @Delta2 posted while I was composing this.
I see it much clearer now!

Then we agree that the potential difference will be the same in both hemispheres and, following my reasoning raised at the beginning, the capacitance would be:

##2\pi \epsilon _0(1+K)(\frac {r_ar_b}{r_b-r_a})##​

??
 
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  • #12
madafo3435 said:
I see it much clearer now!

Then we agree that the potential difference will be the same in both hemispheres and, following my reasoning raised at the beginning, the capacitance would be:

##2\pi \epsilon _0(1+K)(\frac {r_ar_b}{r_b-r_a})##​

??
Yes, we agree. Why the question marks? Isn't the capacitance what you were looking for?
 
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  • #13
kuruman said:
Yes, we agree. Why the question marks? Isn't the capacitance what you were looking for?
I was wrong writing hahaha :-p

Thank you very much for your help! I also thank Delta2 for your contributions.
 
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Related to Find the capacitance of a half-full capacitor

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy by accumulating charge on two conductive plates separated by an insulating material.

2. How does a capacitor work?

A capacitor works by creating an electric field between its two plates. When a voltage is applied, electrons gather on one plate, creating a negative charge, while the other plate becomes positively charged. This creates a potential difference, or voltage, between the plates, and the capacitor stores energy in the form of this electric field.

3. What is the capacitance of a half-full capacitor?

The capacitance of a half-full capacitor is the measure of its ability to store charge. It is directly proportional to the surface area of the plates and inversely proportional to the distance between them. In a half-full capacitor, the capacitance is reduced by half compared to a fully charged capacitor.

4. How do you calculate the capacitance of a half-full capacitor?

The capacitance of a half-full capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the insulating material, A is the surface area of the plates, and d is the distance between the plates. This formula assumes that the plates are parallel and have equal surface areas.

5. What factors affect the capacitance of a half-full capacitor?

The capacitance of a half-full capacitor is affected by the surface area of the plates, the distance between the plates, and the permittivity of the insulating material. It is also affected by the dielectric material between the plates, the shape of the plates, and the presence of any other conductive materials nearby that may alter the electric field.

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