Question

A dielectric of relative permittivity (dielectric constant) k completely fills the space between the plates of a parallel-plate capacitor with a surface charge density σ. Show that the induced density of surface charge on the dielectric is σ_p = σ_p (1 – \(\cfrac 1k\))

Answer 1

Consider a parallel-plate capacitor without a dielectric, of plate area A, plate separation d and capacitance C₀, charged to a potential difference V and then isolated.

Suppose the charges on its conducting plates are +Q and -Q, from figure (a). The surface density of free charge is

σ = \(\cfrac QA\)........(1)

A parallel plate capacitor (a) without dielectric (b) with dielectric

When no dielectric is present, the electric field \(\vec{E_0}\) n the region between the plates can be found by applying Gauss’s law to the Gaussian surface :

When a dielectric is inserted as in figure (b), there is an induced charge – Q, on the surface, and . the net charge enclosed by the Gaussian surface is Q – Q_p. Then, by Gauss’s law,

The effect of the dielectric is to weaken the original field E₀ by a factor k.
Writing E = \(\cfrac {E_0}k\).

\(\therefore\) The surface density of induced charge is

σ_p = σ - \(\cfrac σk\) = σ \((1-\cfrac 1k)\).......(4)