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Differential Form of Gauss Law

Differential Form of Gauss Law :- The Differential Form of Gauss’s Law is the local (point-by-point) version of Gauss’s law. Instead of relating the total electric flux through a closed surface to the total enclosed charge, it relates the divergence of the electric field at a point to the charge density at that point.

Gauss’s law in integral form states that

$\displaystyle \oint{{\vec{E}}}\cdot d\vec{A}=\frac{{{Q}_{\text{enclosed}}}}{{{\varepsilon }_{0}}}$

Here, Q_enclosed represents the total charge enclosed inside the Gaussian surface.

When a dielectric medium is present, the enclosed charge consists of two parts :

Free charge (Q_f) – charge supplied by an external source such as a battery.
Bound charge (Q_b) – charge produced due to the polarization of the dielectric.

Therefore,

$\displaystyle {{Q}_{enclosed}}={{Q}_{f}}+{{Q}_{b}}$

Hence,

$\displaystyle \oint{{\vec{E}}}\cdot d\vec{A}=\frac{{{Q}_{f}}+{{Q}_{b}}}{{{\varepsilon }_{0}}}$

Now, $\displaystyle {{Q}_{f}}=\int_{V}{{{\rho }_{f}}}dV$ and $\displaystyle {{Q}_{b}}=\int_{V}{{{\rho }_{b}}}dV$ . Therefore

$\displaystyle \oint{_{S}\vec{E}}\cdot d\vec{A}=\frac{1}{{{\varepsilon }_{0}}}\int_{V}{({{\rho }_{f}}+{{\rho }_{b}})}dV$ …..(1)

Consider any closed surface (Gaussian surface) enclosing a region of space. For simplicity, let us imagine it to be spherical. The orange arrows represent the electric field lines passing through the surface.

Now, instead of looking only at the surface, suppose we divide the entire enclosed volume into millions of tiny cubes. For a very tiny cube,

Net outward flux = dϕ

Volume = dV

The ratio $\displaystyle \frac{d\phi }{dV}$ represents the net outward flux per unit volume. In the limit as the volume shrinks to zero, this quantity is called the divergence of the vector field.

Imagine the tiny cube is centered at the point P (x, y, z) and its sides are dx, dy, dz.

The dot (•) is the point where we want to know the divergence. The cube is so small that the field changes only a tiny amount across it. At any point, the electric field is given by

$\displaystyle \vec{E}={{E}_{x}}(x,y,z)\hat{i}+{{E}_{y}}(x,y,z)\hat{j}+{{E}_{z}}(x,y,z)\hat{k}$

Let us calculate the flux through the two x-faces. Each face has area dydx.

Flux through the right face :

$\displaystyle d{{\phi }_{right}}={{\vec{E}}_{x}}\cdot d{{\vec{A}}_{right}}=\left[ {{E}_{x}}\left( x+\frac{dx}{2} \right)\hat{i} \right]\cdot \left[ dydz\hat{i} \right]={{E}_{x}}\left( x+\frac{dx}{2} \right)dydz$

Flux through the left face :

$\displaystyle d{{\phi }_{left}}={{\vec{E}}_{x}}.d{{\vec{A}}_{left}}=\left[ {{E}_{x}}\left( x-\frac{dx}{2} \right)\hat{i} \right].\left[ dydz(-\hat{i}) \right]=-{{E}_{x}}\left( x-\frac{dx}{2} \right)dydz$

Net x-flux

$\displaystyle d{{\phi }_{x}}={{E}_{x}}\left( x+\frac{dx}{2} \right)dydz-{{E}_{x}}\left( x-\frac{dx}{2} \right)dydz=\left[ {{E}_{x}}\left( x+\frac{dx}{2} \right)-{{E}_{x}}\left( x-\frac{dx}{2} \right) \right]dydz$

Now look at the difference $\displaystyle \left[ {{E}_{x}}\left( x+\frac{dx}{2} \right)-{{E}_{x}}\left( x-\frac{dx}{2} \right) \right]$ , It looks exactly like the numerator of a derivative :

$\displaystyle \frac{\partial {{E}_{x}}}{\partial x}=\underset{dx\to 0}{\mathop{lim}}\,\frac{{{E}_{x}}\left( x+\frac{dx}{2} \right)-{{E}_{x}}\left( x-\frac{dx}{2} \right)}{dx}$

Rearranging this gives,

$\displaystyle {{E}_{x}}\left( x+\frac{dx}{2} \right)-{{E}_{x}}\left( x-\frac{dx}{2} \right)\approx \frac{\partial {{E}_{x}}}{\partial x}dx$

Substitute this into the flux :

$\displaystyle d{{\phi }_{x}}=\frac{\partial {{E}_{x}}}{\partial x}dxdydz$

Similarly for y and z – faces,

$\displaystyle d{{\phi }_{y}}=\frac{\partial {{E}_{y}}}{\partial y}dxdydz$

$\displaystyle d{{\phi }_{z}}=\frac{\partial {{E}_{z}}}{\partial z}dxdydz$

Total flux passing through all three faces,

$\displaystyle d\phi =\left( \frac{\partial {{E}_{x}}}{\partial x}+\frac{\partial {{E}_{y}}}{\partial y}+\frac{\partial {{E}_{z}}}{\partial z} \right)dxdydz$

But $\displaystyle dxdydz=dV$ , so

$\displaystyle d\phi =\left( \frac{\partial {{E}_{x}}}{\partial x}+\frac{\partial {{E}_{y}}}{\partial y}+\frac{\partial {{E}_{z}}}{\partial z} \right)dV$

$\displaystyle \Rightarrow \frac{d\phi }{dV}=\left( \frac{\partial {{E}_{x}}}{\partial x}+\frac{\partial {{E}_{y}}}{\partial y}+\frac{\partial {{E}_{z}}}{\partial z} \right)$

We know that the differential operator

$\displaystyle \vec{\nabla }=\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}$

Taking its dot product with $\displaystyle \vec{E}={{E}_{x}}\hat{i}+{{E}_{y}}\hat{j}+{{E}_{z}}\hat{k}$ , gives

$\displaystyle \vec{\nabla }\cdot \vec{E}=\left( \hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z} \right)\cdot \left( {{E}_{x}}\hat{i}+{{E}_{y}}\hat{j}+{{E}_{z}}\hat{k} \right)=\frac{\partial {{E}_{x}}}{\partial x}+\frac{\partial {{E}_{y}}}{\partial y}+\frac{\partial {{E}_{z}}}{\partial z}$

Hence,

$\displaystyle \vec{\nabla }\cdot \vec{E}=\frac{d\phi }{dV}$ …..(2)

Now total electric flux passing through a closed surface,

$\displaystyle \oint\limits_{S}{{\vec{E}}}\cdot d\vec{A}=\int\limits_{V}{d\phi }$ …..(3)

From equation (2)

$\displaystyle d\phi =\left( \vec{\nabla }\cdot \vec{E} \right)dV$

Using this in equation (3)

$\displaystyle \oint\limits_{S}{{\vec{E}}}\cdot d\vec{A}=\int\limits_{V}{\left( \vec{\nabla }\cdot \vec{E} \right)dV}$ …..(4)

The above result is the Divergence Theorem, which states that :

The total outward flux of a continuously differentiable vector field through a closed surface is equal to the volume integral of the divergence of that field over the region enclosed by the surface.

Symbolically,

(Total outward flow through surface) = (Sum of local outward flow inside)

Finally, since both equations (1) and (4) represent the total electric flux through the same closed surface, we obtain

$\displaystyle \int\limits_{V}{(\nabla \cdot \vec{E})}dV=\frac{1}{{{\varepsilon }_{0}}}\int\limits_{V}{\rho }dV$

$\displaystyle \Rightarrow \int\limits_{V}{\left( \nabla \cdot \vec{E}-\frac{\rho }{{{\varepsilon }_{0}}} \right)}dV=0$

Since this equation is true for every arbitrary closed volume, the quantity inside the integral must be zero at every point in space. Hence,

$\displaystyle \nabla \cdot \vec{E}-\frac{\rho }{{{\varepsilon }_{0}}}=0$

$\displaystyle \Rightarrow \nabla \cdot \vec{E}=\frac{\rho }{{{\varepsilon }_{0}}}$

Hence

$\displaystyle \nabla \cdot \vec{E}=\frac{\rho }{{{\varepsilon }_{0}}}=\frac{{{\rho }_{f}}+{{\rho }_{b}}}{{{\varepsilon }_{0}}}$ …..(5)

This is what we call Differential Form of Gauss’s Law.

Differential Form of Gauss Law

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