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Question

A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the xy plane, centered at the origin. The top plate is located at z = d, with its center on the z axis. Potential V0V_0 is on the top plate; the bottom plate is grounded. Dielectric having radially dependent permittivity fills the region between plates. The permittivity is given by ϵ(ρ)=ϵ0(1+ρ2/a2)\epsilon(\rho)=\epsilon_{0}\left(1+\rho^{2} / a^{2}\right) Find (a)V(z); (b) E; (c) Q; (d)C.

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a)\textbf{a)}

We have learned in the Problem 6.36. that even if the permittivity varies in space, the Poisson equation is still valid if permittivity varies in a direction perpendicular to the electric field. That is the case here, so we can start with the Laplace equation (since ρv=0\rho_v=0):

2V=0    2Vz2=0\begin{gather*} \nabla^2V=0\implies \dfrac{\partial^2V}{\partial z^2}=0 \end{gather*}

The solution that satifies the boundary conditions is a simple linear function:

V(z)=V0zd\begin{gather*} \boxed{V(z)=V_0\dfrac{z}{d}} \end{gather*}

b)\textbf{b)}

The electric field is the negative gradient of the potential:

E=V=V0daz\begin{gather*} \mathbf E=-\nabla V=-\boxed{\dfrac{V_0}{d}\mathbf a_z} \end{gather*}

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