E LECTRIC DISPLACEMENT We have introduced the notion of dielectric constant and arrived at Eq. ( . ), without giving the explicit relation between the induced charge density σ p and the polarisation P . We take without proof the result that P ˆ σ = P n where ˆ n is a unit vector along the outward normal to the surface.
Above equation is general, true for any shape of the dielectric. For the slab in Fig. . , P is along ˆ n at the right surface and opposite to ˆ n at the left surface.
Thus at the right surface, induced charge density is positive and at the left surface, it is negative, as guessed already in our qualitative discussion before. Putting the equation for electric field in vector form ˆ ˆ σ P n E n or ( ε E + P ) ˆ n = σ The quantity ε E + P is called the electric displacement and is denoted by D . It is a vector quantity. Thus, D = ε E + P , D ˆ n = σ, The significance of D is this : in vacuum, E is related to the free charge density σ .
When a dielectric medium is present, the corresponding role is taken up by D . For a dielectric medium, it is D not E that is directly related to free charge density σ , as seen in above equation. Since P is in the same direction as E , all the three vectors P , E and D are parallel. The ratio of the magnitudes of D and E is P D K E σε σ σ Thus, D = ε K E and P = D – ε E = ε ( K – ) E This gives for the electric susceptibility χ e defined in Eq.
( . ) χ e = ε ( K – ) Example . A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness ( / ) d , where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
Solution Let E = V / d be the electric field between the plates when there is no dielectric and the potential difference is V . If the dielectric is now inserted, the electric field in the dielectric will be E = E / K . The potential difference will then be E XAMPLE . ( ( E E K ( K E d K K The potential difference decreases by the factor ( K + )/ K while the free charge Q on the plates remains unchanged.
The capacitance thus increases Q Q K K K K