Gauss Law in Dielectrics

Consider a parallel-plate capacitor with pate area A and having vacuum between its plates. Let + q and - q be the charges on the plates and E0 the uniform electric field between the plates. Let PQRS be a Gaussian surface.

By Gauss’s law, φ E₀. dS = q/ε₀

E0 || dS and constant over the Guassian surface.
    Φ E₀. dS = Φ E₀dS = E₀A
. :    E₀A = q/ε₀ pr E₀ = q/ε₀A.                                … (1)

Suppose a material of dielectric constant K is introduced in the intervening space between the two plates. The dielectric slab gets polarized. A negative charged – q’ and + q’ on the dielectric are called the ‘induced charges’ or ‘bound charges’ while the charges +q and – q on the capacitor plates are called free charges. These induced charges produce their own field which opposes the external field E₀. Let E be the resultant field within the dielectric. The net charge within the Gaussian surface is q – q’.

. : by Gauss’s Law, Φ E. dS = q – q’/ε₀                             … (2)

Or EA = q – q’/ε₀     or     E = = q – q’/ε₀ A                        … (3)

Now, E₀/E = K, where K is dielectric constant.
Eq. (1) becomes, E = q/ Kε₀A
Inserting this in Eq. (3), q/Kε₀A = q – q’/ε₀A

Or            q – q’ = q/K
Substituting this value of q – q’ in Eq. (2),
            Φ E. dS = q/Kε₀   

. :            Φ k E. dS = q/ε₀                             … (4)

It is the Gauss’s law in the presence of a dielectric. Here we see that the flux integral contains a factor K. the effect of the induce surface charge is ignored by taking into accounring K, the dielectric contant.
For dielectric constant, εr is used instead of K.



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