## Gauss Law In Dielectrics

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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 E

_{0}. dS = q/ε

_{0}

E0 || dS and constant over the Guassian surface.

Φ E

_{0}. dS = Φ E

_{0}dS = E

_{0}A

. : E

_{0}A = q/ε

_{0}pr E

_{0}= q/ε

_{0}A. … (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 E

_{0}. 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’/ε

_{0}… (2)

Or EA = q – q’/ε

_{0}or E = = q – q’/ε

_{0 }A … (3)

Now, E

_{0}/E = K, where K is dielectric constant.

Eq. (1) becomes, E = q/ Kε

_{0}A

Inserting this in Eq. (3), q/Kε

_{0}A = q – q’/ε

_{0}A

Or q – q’ = q/K

Substituting this value of q – q’ in Eq. (2),

Φ E. dS = q/Kε

_{0}

. : Φ k E. dS = q/ε

_{0}… (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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