Question

# The gap between the plates of a parallel-plate capacitor is filled up with two dielectric layers $$1$$ and $$2$$ with thicknesses $$d_{1}$$ and $$d_{2}$$, permittivities $$\epsilon_{1}$$ and $$\epsilon_{2}$$, and resistivities $$\rho_{1}$$ and $$\rho_{2}$$. A dc voltage $$V$$ is applied to the capacitor, with electric field directed from layer $$1$$ to layer $$2$$. Find $$\sigma$$, the surface density of extraneous charges at the boundary between the dielectric layers, and the condition under which $$\sigma = 0$$.

Solution

## We have, $$E_{1}d_{1} + E_{2}d_{2} = V$$and by current conservation$$\dfrac {1}{\rho_{1}} E_{1} = \dfrac {1}{\rho_{2}}E_{2}$$Thus, $$E_{1} = \dfrac {\rho_{1}V}{\rho_{1}d_{1} + \rho_{2}d_{2}}$$,$$E_{2} = \dfrac {\rho_{2}V}{\rho_{1}d_{1} + \rho_{2}d_{2}}$$At the boundary between the two dielectrics,$$\sigma = D_{2} - D_{1} = \epsilon_{0} \epsilon_{2} E_{2} - \epsilon_{0} \epsilon_{1} E_{1}$$$$\dfrac {\epsilon_{0}V}{\rho_{1}d_{1} + \rho_{2}d_{2}} (\epsilon_{2} \rho_{2} - \epsilon_{1} \rho_{1})$$.Physics

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