# Capacitor and Surface Charge Density Question

• cwill53
In summary: V }{\partial z^2}=0$$because, under ideal conditions, there is only a change in potential in one direction of the parallel plate. cwill53 Homework Statement A capacitor consists of a pair of large parallel metal plates 3.0 mm apart, between which lies a parallel sheet of mica 2.5 mm thick; air fills the rest of the space between the plates. If the mica be taken to have a permittivity of 5, find the density of charge on the plates, and the voltage gradient in the air, when the capacitor is charged to a potential difference of 2000 V. If the mica is replaced by a pair of thin parallel metal sheets separated by 2.5 mm of air, find the density of charge needed on these sheets in order that the charge on the outer plates shall be the same as when the mica was in place; the potential difference of the capacitor being kept constant at 2000 V. Answers: ##17.6\cdot 10^{-6} C/m^2;2.0MV/m;14.2\cdot 10^{-6}C/m^2## Relevant Equations$$C=\frac{Q}{V}=\frac{Q}{Q-q}\frac{\varepsilon _0S}{d}=\varepsilon_r\frac{\varepsilon_0S}{d}\Rightarrow \frac{Q}{S}=\frac{V\varepsilon_0\varepsilon_r}{d}=\frac{V\varepsilon}{d}$$When I plug in the numbers I get ##2.9513\cdot 10^{-5}C/m^2##, not ##17.6\cdot 10^{-6} C/m^2##. Can someone point out where I'm going wrong? It seems that you are trying to find the capacitance with mica in. I do not understand how you go from ##C=\dfrac{Q}{V}## to ##\dfrac{Q}{Q-q}\dfrac{\epsilon_0S}{d}##. What is ##q## and why does it go away? Can you explain? I would find the capacitance by considering two capacitors in series, one with mica in the space between its plates and one with air between its plates. cwill53 kuruman said: It seems that you are trying to find the capacitance with mica in. I do not understand how you go from ##C=\dfrac{Q}{V}## to ##\dfrac{Q}{Q-q}\dfrac{\epsilon_0S}{d}##. What is ##q## and why does it go away? Can you explain? I would find the capacitance by considering two capacitors in series, one with mica in the space between its plates and one with air between its plates. ##q## is the induced charge on the insulator between the plates, in this case, the mica. The ratio ##\dfrac{Q}{Q-q}## is the relative permittivity of the mica. The right answer can be found by considering two capacitors in series, but I think there's a better way unless we specifically say:"consider this as two capacitors in series" without further explanation. Gordianus said: The right answer can be found by considering two capacitors in series, but I think there's a better way unless we specifically say:"consider this as two capacitors in series" without further explanation. I would’ve done it that way if the problem statement had’ve expressed it like that. I was under the assumption that I was supposed to take it as just one capacitor with some dielectric in between. cwill53 said: I would’ve done it that way if the problem statement had’ve expressed it like that. I was under the assumption that I was supposed to take it as just one capacitor with some dielectric in between. It's one capacitor with some dielectric and some air in between that is exactly equivalent to a capacitor with a gap of 2.5 mm filled with mica in series with an air-filled capacitor with a gap of 0.5 mm. The most direct approach at the introductory physics level is the two capacitors in series route. You can, of course, solve Laplace's equation and match boundary conditions, but that would be overkill. Nevertheless, by not specifying how to go about it, the author of the problem left it up to you. kuruman said: It's one capacitor with some dielectric and some air in between that is exactly equivalent to a capacitor with a gap of 2.5 mm filled with mica in series with an air-filled capacitor with a gap of 0.5 mm. The most direct approach at the introductory physics level is the two capacitors in series route. You can, of course, solve Laplace's equation and match boundary conditions, but that would be overkill. Nevertheless, by not specifying how to go about it, the author of the problem left it up to you. Can you walk me through how to solve Laplace's equation for this problem? I have some experience with Fourier series and separation of variables. Delta2 cwill53 said: Can you walk me through how to solve Laplace's equation for this problem? I have some experience with Fourier series and separation of variables. Sure. Start by writing Laplace's equation in Cartesian coordinates. Adapt it to this particular problem, noting that the plates are "large" and that dimensions are given in only one direction. That could mean only one thing. Solve the ensuing equation and see what you get. No Fourier series or separation of variables needed. Once you've done all that, we'll take it from there. cwill53 and Delta2 If we want to resort to Laplace's equation, we need to split the problem into two parts because we have a polarization surface charge at the interface between mica and air. cwill53 and Delta2 Gordianus said: If we want to resort to Laplace's equation, we need to split the problem into two parts because we have a polarization surface charge at the interface between mica and air. We need to split the region between the plates into region I (with mica) and region II (with air). Continuity of the normal component of ##\vec D## at the interface takes care of the surface polarization. Delta2 kuruman said: Sure. Start by writing Laplace's equation in Cartesian coordinates. Adapt it to this particular problem, noting that the plates are "large" and that dimensions are given in only one direction. That could mean only one thing. Solve the ensuing equation and see what you get. No Fourier series or separation of variables needed. Once you've done all that, we'll take it from there.$$\nabla^2V=\frac{\partial^2V }{\partial x^2}+\frac{\partial^2V }{\partial y^2}+\frac{\partial^2V }{\partial z^2}=0$$I assume this reduces to$$\frac{\partial^2V }{\partial z^2}=0$$because, under ideal conditions is only a change in potential in one direction of the parallel plate capacitor. Right. Now get rid of the partial differentials because it is a one-dimensional problem and solve the diff. eq. Note that each of the two regions I and II has its own general solution with its own integration constants. kuruman said: We need to split the region between the plates into region I (with mica) and region II (with air). Continuity of the normal component of ##\vec D## at the interface takes care of the surface polarization. ##\vec D## is what exactly? Is this the flux density? After solving, I end up getting, in general, that$$V(z)=C_1z+C_2$$I'm thinking now I would have to set the right boundary conditions to find the constants. cwill53 said: ##\vec D## is what exactly? Is this the flux density? After solving, I end up getting, in general, that$$V(z)=C_1z+C_2$$I'm thinking now I would have to set the right boundary conditions to find the constants. Like I said kuruman said: Note that each of the two regions I and II has its own general solution with its own integration constants. You need to write two potentials, e.g.$$\begin{align} & V_{\text{I}}(z)=B_1z+B_2\nonumber \\ & V_{\text{II}}(z)=C_1z+C_2 \nonumber\end{align}$$Vector ##\vec D## is the electric displacement field. In most cases it is related to the electric field by ##\vec D=\epsilon \vec E##. Gauss's law in the presence of dielectrics takes the form as$$\int_S \vec D\cdot \hat n~dA=\int_V \rho~dV$$where ##\rho## is the free charge density. kuruman said: Like I said You need to write two potentials, e.g.$$\begin{align} & V_{\text{I}}(z)=B_1z+B_2\nonumber \\ & V_{\text{II}}(z)=C_1z+C_2 \nonumber\end{align}$$Vector ##\vec D## is the electric displacement field. In most cases it is related to the electric field by ##\vec D=\epsilon \vec E##. Gauss's law in the presence of dielectrics takes the form as$$\int_S \vec D\cdot \hat n~dA=\int_V \rho~dVwhere ##\rho## is the free charge density.
What would be the appropriate boundary conditions for each region? I see we have a potential for each region, but I'm confused as to how we account for the mica between the plates in the boundary conditions. I know we're splitting the region into two, one air and one mica, do we just say that z varies from 0 to 2.5 millimeters for the mica region and z varies from 2.5 mm to 3 mm for the air region?

There are three boundaries. One is plate I, one is plate II and one is the dielectric-air interface.
You need to set the values of the potential at each plate. Also, the potential must be continuous at the interface.
The normal component of ##\vec D## is discontinuous at the plates because the electric field is zero inside the plates. The discontinuity of the normal component of ##\vec D## is equal to the surface charge density.

To account for the mica, note that the normal component of ##\vec D## must be continuous at the interface. You know that ##\vec E\cdot \hat n =-\dfrac{dV}{dz}## and that ##\vec D=\epsilon \vec E## so ##\dots##

There is a lot of how-to information in this post. Please study it carefully and understand what I am saying before jumping in headfirst.

cwill53

## 1. What is a capacitor?

A capacitor is an electronic component that is used to store electrical energy in an electric field. It is made up of two conductive plates separated by an insulating material, known as a dielectric. When a voltage is applied to the capacitor, it stores electric charge on its plates.

## 2. How is the capacitance of a capacitor calculated?

The capacitance of a capacitor is calculated by dividing the charge stored on its plates by the voltage applied to it. It is also affected by the distance between the plates, the surface area of the plates, and the type of dielectric material used.

## 3. What is surface charge density?

Surface charge density is the amount of electric charge per unit area on the surface of a conductor. In the case of a capacitor, it is the amount of charge per unit area on the surface of its plates.

## 4. How does surface charge density affect the capacitance of a capacitor?

The surface charge density on the plates of a capacitor is directly proportional to the capacitance. This means that as the surface charge density increases, the capacitance also increases. This is because a higher surface charge density means there is more charge stored on the plates, resulting in a larger capacitance.

## 5. How does the dielectric material affect the surface charge density of a capacitor?

The dielectric material used in a capacitor affects the surface charge density by influencing the electric field between the plates. A higher dielectric constant, which measures the ability of a material to store electric charge, results in a higher surface charge density and therefore a higher capacitance.

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