We have a parallel plate capacitor and provide a potential V across, leading to +Q and -Q charges on the respective plates. Now, in this condition, we add a sheet between the plates that have positive charges stuck on the sheet. The motive for adding this positively charged sheet between the plates is to neutralize the field between the capacitor plates; hence after adding this positively charged sheet, the plate with +Q charge will feel no net electric field. This is because the electric field that was initially between the plates is now equal and opposite to the electric field from Charged sheet. So, the question is, what will happen to the charge on the Plate (which initially had charge +Q on it)? Will the charge on plate become zero?(NOTE: we are still providing V potential) Intial condition
Final after adding positively charged plate which neutralizes Electric field felt by plate that had +Q charge initially


1 Answer 1


So this problem can be solved by solving the equation $\text{div}\mathbf{D}=\rho$ with the electric scalar potential $\phi$ as an unknown.

Assume the capacitor is in the range $[0,L]$ along the x-axis. The width of the charged region is $\Delta$ and its center coordinate is $X_0$. The uniform volume charge density inside the region is $\rho_1$ in unit of $C/m^3$. The region with the electric charge distribution should be an insulator and the dielectric constant is $\epsilon_1$. The left and right electrodes are assumed to be scalar potential-fixed Dirichlet boundary conditions. enter image description here

Up to now the parameters $L, V, X_0, \Delta, \epsilon_0, \epsilon_1$ and $\rho_1$ have been assumed to be given constants. In the following, 4 unknown parameters are included; $E_{\text{left}}$ is the electric field in the left region, $E_{\text{right}}$ is the electric field in the right region, a and b unknown constants.

The electric scalar potential sholud be of the form as, \begin{equation*} \phi(x)= \begin{cases} -E_{\text{left}}x+V \text{ for left}\\ -\frac{1}{2}\frac{\rho_1}{\epsilon_1}x^2-ax+b \text{ for charged} \\ -E_{\text{right}}(x-L) \text{ for right}. \end{cases} \end{equation*} The electric field is calculated as, \begin{equation*} E_x(x)=-\frac{\partial\phi(x)}{\partial x}= \begin{cases} E_{\text{left}} \text{ for left} \\ \frac{\rho_1}{\epsilon_1}x+a \text{ for charged} \\ E_{\text{right}}\text{ for right} \end{cases} \end{equation*} Finally, the electric density function is calculated as, \begin{equation*} D_x(x)=\epsilon(x)E_x(x)=\epsilon(x)\left(-\frac{\partial\phi(x)}{\partial x}\right)= \begin{cases} \epsilon_0E_{\text{left}} \text{ for left}\\ \epsilon_1\left(\frac{\rho_1}{\epsilon_1}x+a\right) \text{ for charged}\\ \epsilon_0E_{\text{right}}\text{ for right}. \end{cases} \end{equation*}

We impose some continuos conditions at $x=\left(X_0-\frac{\Delta}{2}\right)$ and $x=\left(X_0+\frac{\Delta}{2}\right)$: \begin{align*} -E_{\text{left}}\left(X_0-\frac{\Delta}{2}\right)+V=-\frac{1}{2}\frac{\rho_1}{\epsilon_1}\left(X_0-\frac{\Delta}{2}\right)^2-a\left(X_0-\frac{\Delta}{2}\right)+b \tag{1}\\ -\frac{1}{2}\frac{\rho_1}{\epsilon_1}\left(X_0+\frac{\Delta}{2}\right)^2-a\left(X_0+\frac{\Delta}{2}\right)+b =-E_{\text{right}}\left(X_0+\frac{\Delta}{2}-L\right) \tag{2}\\ \epsilon_0E_{\text{left}}=\epsilon_1\left(\frac{\rho_1}{\epsilon_1}\left(X_0-\frac{\Delta}{2}\right)+a\right) \tag{3}\\ \epsilon_1\left(\frac{\rho_1}{\epsilon_1}\left(X_0+\frac{\Delta}{2}\right)+a\right)=\epsilon_0E_{\text{right}} \tag{4} \end{align*} The first 2 equations, (1) and (2), are the continuous conditions for $\phi(x)$ and the last 2 equations, (3) and (4) are the continuous conditions for $D_x(x)$. Note that the electric field $E_x(x)$ is not continuous.

The equations (1) to (4) are linear simultaneous equations, so they can be solved analytically. But I'm a bit tired, gave up on getting analytic forms.

Is $E_\text{left}=0$ possible?

Let us check if \begin{equation*} E_\text{left}=0 \tag{5} \end{equation*} possible or not. From (3) and (4), we get, \begin{equation*} \epsilon_0\left(E_{\text{right}}-E_{\text{left}}\right)=\rho_1\Delta \tag{6} \end{equation*} Substituting (5) into (6) gives \begin{equation*} E_{\text{right}}=\frac{\rho_1}{\epsilon_0}\Delta \tag{7} \end{equation*} Substituting (5) into (3) gives \begin{equation*} a=-\frac{\rho_1}{\epsilon_1}\left(X_0-\frac{\Delta}{2}\right) \tag{8} \end{equation*} And substituting (5), (7) and (8) into (1) gives \begin{equation*} b=V-\frac{1}{2}\frac{\rho_1}{\epsilon_1}\left(X_0-\frac{\Delta}{2}\right)^2 \tag{9}. \end{equation*} Finally, if we insert (5), (7), (8) and (9) into (2), we get \begin{equation*} \rho_1\Delta = \frac{\epsilon_0V}{L-\left(X_0+\frac{\Delta}{2}\right)+\frac{\epsilon_0}{\epsilon_1}\frac{\Delta}{2}}. \tag{10} \end{equation*} Given the specific electric charge distribution represented by this equation (10), we can set $E_\text{left}=0$.


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