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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.

Gauss Law in Dielectrics

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