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) E; (b) D; (c) Q; (d) C.

Solution

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a)\textbf{a)}

The electric potential is still going to be V(z)=V0z/dV(z)=V_0z/d, so the electric field, being the negative gradient of the potential, is:

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

b)\textbf{b)}

The displacement field is:

D=ϵE=ϵ0(1+ρ2a2)V0daz\begin{gather*} \mathbf D=\epsilon \mathbf E=\boxed{-\epsilon_0\left(1+\dfrac{\rho^2}{a^2}\right)\dfrac{V_0}{d}\mathbf a_z} \end{gather*}

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