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Prove $\nabla .\bar{D} = \rho_v$

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Prove $\nabla .\bar{D} = \rho_v$

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written 6.9 years ago by |

According to Gauss’s law for electric field, the total flux crossing the closed surface is equal to the charge enclosed by that surface

The electric flux through the closed surface is,

$ψ= ∮_s \bar{D} .d \bar{s}$

The charge can be expressed in terms of $ρ_v$ as:

$Q_{enclosed}= ∫_v ρ_v.dv$

The gauss law of electric is expressed mathematically as

$∮_s \bar{D}.d \bar{s}= ∫_vρ_v.dv$

To relate $\bar{D}$ with $ρ_v$ , using divergence theorem as

$∮_s \bar{D} .d \bar{s}= ∫_v(∇ .\bar{D} ) dv \\ ∫(∇ .\bar{D} ) dv= ∫_vρ_v dv \\ ∇ .\bar{D}=ρ_v$

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