How can I relate the electric field to potential in a parallel plate capacitor?

In summary: I do not know what does the term "legendary polynomial" mean?It's "Legendre polynomials". It is the name of the person who invented that particular type of math. A large number of mathematical (and scientific) concepts bear the name of the person who invented the concept.thanks.
  • #1
Artyman
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I want to calculate Electric field inside and outside a capacitor(two parallel conductor plates and in between a dielectric) which is fed by an AC voltage source. My problem is I do not know how can I relate the electric field to the potential in a capacitor. I got confused about the derivative of V respect to distance...
Assuming: the parallel plates are in a distance in "X" direction in cartesian coordinate. But there is no factor of X in the potential equation to calculate the electric field in any distance between two plates.
 
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  • #2
Is this a homework problem? If so, please use the homework template.
Artyman said:
My problem is I do not know how can I relate the electric field to the potential in a capacitor. I got confused about the derivative of V respect to distance...
The relationship between ##\mathbf{E}## and the potential ##V(x)## in the x-direction is
$$E_{x}=-\frac{\partial V(x)}{\partial x}$$
Since ##\mathbf{E}## is constant between the plates, what must the function ##V(x)## look like?
 
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  • #3
NFuller said:
Is this a homework problem? If so, please use the homework template.

The relationship between ##\mathbf{E}## and the potential ##V(x)## in the x-direction is
$$E_{x}=-\frac{\partial V(x)}{\partial x}$$
Since ##\mathbf{E}## is constant between the plates, what must the function ##V(x)## look like?
just V(x)=X*(constant)? it means even for a capacitor in an AC circuit the Electric field is constant for all the point between the plates, but potential is a linear function of distance?
Thanks
 
  • #4
Artyman said:
just V(x)=X*(constant)?
Which constant?
Artyman said:
it means even for a capacitor in an AC circuit the Electric field is constant for all the point between the plates, but potential is a linear function of distance?
Yes. The field is constant w.r.t. distance (uniform field) but varies w.r.t. time.
 
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  • #5
Artyman said:
just V(x)=X*(constant)? it means even for a capacitor in an AC circuit the Electric field is constant for all the point between the plates, but potential is a linear function of distance?
Right. Now you need to find the constant which makes the function ##V(x)## match the potential at the plates.
 
  • #6
cnh1995 said:
Which constant?

Yes. The field is constant w.r.t. distance (uniform field) but varies w.r.t. time.
The two metal plates are circular, parallel and in 3Dimentions are located on the surface of a sphere, the metal plates are connected to AC source but low frequency.
we assume the sphere is solid and is made of a (dielectric)material with conductivity and permittivity (sigma, epsilon). So, now the Electric field is constant in all the points in between and out of the plates?
thanks
 
  • #7
Artyman said:
The two metal plates are circular, parallel and in 3Dimentions are located on the surface of a sphere, the metal plates are connected to AC source but low frequency.
we assume the sphere is solid and is made of a (dielectric)material with conductivity and permittivity (sigma, epsilon). So, now the Electric field is constant in all the points in between and out of the plates?
thanks
Is this a different question? If so please start a new thread.
 
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  • #8
NFuller said:
Is this a different question? If so please start a new thread.
No, just the details about my question. first I made it just simplified.
 
  • #9
Okay, for a spherical linear dielectric the electric potential will be composed of Legendre polynomials. Do you know the general solution to the Laplace equation in spherical coordinates?
 
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  • #10
NFuller said:
Okay, for a spherical linear dielectric the electric potential will be composed of Legendre polynomials. Do you know the general solution to the Laplace equation in spherical coordinates?
sorry I do not know what does the term "legendary polynomial" mean?
 
  • #11
Maybe I misunderstood the description of the set-up. Does it look like this
upload_2017-9-10_18-57-46.png

or like this
upload_2017-9-10_19-2-53.png
 
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  • #12
Artyman said:
sorry I do not know what does the term "legendary polynomial" mean?

It's "Legendre polynomials". It is the name of the person who invented that particular type of math. A large number of mathematical (and scientific) concepts bear the name of the person who invented the concept.

BoB
 
  • #13
NFuller said:
Maybe I misunderstood the description of the set-up. Does it look like this
View attachment 210804
or like this
View attachment 210805
sphere.jpg

Thanks.
I wanted to calculate the E-field at each point in the sphere. The two contacts on the surface of the sphere ( tangent to it).
 
  • #14
I think you can use the potential of two opposite point charges to begin. The issue will be finding the potential which meets the boundary conditions at the surface of the dielectric sphere. To do this, you could use the method of images.

I should mention that without a strong math background this problem will be very intractable.
 
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  • #15
NFuller said:
I think you can use the potential of two opposite point charges to begin. The issue will be finding the potential which meets the boundary conditions at the surface of the dielectric sphere. To do this, you could use the method of images.

I should mention that without a strong math background this problem will be very intractable.
Thank you.
For boundary condition can you guide me which factors do I have to consider?
 
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  • #16
If we call the electric field inside the sphere ##\mathbf{E}_{in}## and the electric field outside ##\mathbf{E}_{out}## then at the surface of the sphere the following must hold:
$$(\epsilon_{0}\mathbf{E}_{out}-\epsilon\mathbf{E}_{in})\cdot\hat{r}=0$$
$$(\mathbf{E}_{out}-\mathbf{E}_{in})\times\hat{r}=0$$
 
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  • #17
NFuller said:
If we call the electric field inside the sphere ##\mathbf{E}_{in}## and the electric field outside ##\mathbf{E}_{out}## then at the surface of the sphere the following must hold:
$$(\epsilon_{0}\mathbf{E}_{out}-\epsilon\mathbf{E}_{in})\cdot\hat{r}=0$$
$$(\mathbf{E}_{out}-\mathbf{E}_{in})\times\hat{r}=0$$
Thank you. So, the only difference is not the dielectric constant in two mediums? the directions of the electric filed will be spherical and pointing the r. Am I right?
 
  • #18
Artyman said:
the directions of the electric filed will be spherical and pointing the r. Am I right?
No, there can be nonzero components perpendicular to ##\hat{r}## which satisfy both equations. For this problem the field will have more than just ##\hat{r}## dependence.
 
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  • #19
NFuller said:
No, there can be nonzero components perpendicular to ##\hat{r}## which satisfy both equations. For this problem the field will have more than just ##\hat{r}## dependence.
thank you. is there any reference or paper that you can refer to me to get the idea?
 
  • #20
Yes, this comes from Classical Electrodynamics by Jackson in chapter 4. This book goes into great detail on these types of problems as well as many others. It is a graduate level text though so it assumes a strong background in math. These topics are also covered in Introduction to Electrodynamics by Griffiths. This text is much more tractable than the Jackson text for beginning E&M, but it is not as detailed.
 
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  • #21
NFuller said:
Yes, this comes from Classical Electrodynamics by Jackson in chapter 4. This book goes into great detail on these types of problems as well as many others. It is a graduate level text though so it assumes a strong background in math. These topics are also covered in Introduction to Electrodynamics by Griffiths. This text is much more tractable than the Jackson text for beginning E&M, but it is not as detailed.
really appreciate it. If I go through it, can I still ask questions here?
 
  • #22
Artyman said:
really appreciate it. If I go through it, can I still ask questions here?
Of course. Good luck!
 
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  • #23
NFuller said:
Of course. Good luck!
thank you.
 

1. What is the relationship between electric field and potential in a parallel plate capacitor?

The electric field and potential in a parallel plate capacitor are directly related. The electric field is the rate of change of potential, or the gradient of potential. This means that as the potential increases or decreases, the electric field also increases or decreases, respectively. In a parallel plate capacitor, the electric field is constant and the potential varies linearly between the plates.

2. How can I calculate the electric field in a parallel plate capacitor?

The electric field in a parallel plate capacitor can be calculated by dividing the voltage difference between the plates by the distance between them. This can be expressed as E = V/d, where E is the electric field, V is the voltage, and d is the distance between the plates.

3. What is the significance of the electric field and potential in a parallel plate capacitor?

The electric field and potential in a parallel plate capacitor play a crucial role in the functioning of electronic devices. The electric field creates a force that moves charges between the plates, while the potential difference allows for the storage and release of electrical energy.

4. Can the electric field and potential in a parallel plate capacitor be manipulated?

Yes, the electric field and potential in a parallel plate capacitor can be manipulated by changing the distance between the plates or the voltage applied to the plates. Increasing the voltage or decreasing the distance between the plates will result in a stronger electric field and higher potential difference.

5. How does the electric field and potential in a parallel plate capacitor change with different materials?

The electric field and potential in a parallel plate capacitor are affected by the dielectric constant of the material between the plates. A higher dielectric constant will result in a lower electric field and potential, while a lower dielectric constant will result in a higher electric field and potential. The material between the plates also affects the capacitance of the capacitor.

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