Question

What is the differential form of Gauss law?

Answer 1

Hint: Electric flux measures the number of electric field lines passing through a point. The electric flux through a surface can be defined as the product of the electric field and the area of the surface through which it passes normally. Gauss’s law is useful in finding the electric charge in a closed surface.

Complete answer:
According to Gauss’s theorem-“electric flux through a closed surface that enclose charge in a vacuum is equal to \[\dfrac{1}{{{\in }_{0}}}\] times the total charge that is enclosed by the surface”.
Gauss law is represented by- \[\oint{\vec{E}.d\vec{s}=\dfrac{q}{{{\in }_{0}}}}\]
Where ‘q’ represents the total charge of the surface and ‘\[{{\in }_{0}}\]’ represents the permittivity of free space.
There are various applications of Gauss’s theorem which are as follows.
The electric field intensity due to an infinitely long charged wire is given by-
\[E=\dfrac{\lambda }{2\pi {{\in }_{0}}r}\],
Where, \[\lambda =\dfrac{q}{l}\]
The electric field intensity due to an plane sheet of charge which is infinite long is given by-
\[E=\dfrac{\sigma }{2{{\in }_{0}}}\]
Where, \[\sigma =\dfrac{q}{A}\]
According to the differential form of Gauss’s law-“The divergence of the electric field at any point in space is equal to \[\dfrac{1}{{{\in }_{0}}}\] times the volume charge density ‘\[\rho \]’ at the point”.
Gauss divergence theorem is represented by \[\nabla .E=\dfrac{\rho }{{{\in }_{0}}}\].
Now we will prove this relation.
We know that, according to Gauss’s theorem
\[\oint{\vec{E}.d\vec{s}=\dfrac{q}{{{\in }_{0}}}}\]……….eq(1)
And we also know that the formula of volume charge density is given by-
\[q=\rho dV\]……………eq(2)
On putting eq(2) in eq(1) and rewriting the right side of eq(1) in terms of volume integral, we get
\[\oint{\vec{E}.d\vec{s}=\oint{\dfrac{\rho }{{{\in }_{0}}}}}dV\]……..eq(3)
According to the gauss divergence theorem-
\[\oint{\vec{E}.d\vec{s}=\oint{(\nabla .E)dV}}\]……..eq(4)
On comparing eq(3) and eq(4), we get
\[\oint{\dfrac{\rho }{{{\in }_{0}}}dV=\oint{(\nabla .E)dV}}\]
Which gives
\[\nabla .E=\dfrac{\rho }{{{\in }_{0}}}\]……eq(5)
Eq(5) represents the differential form of Gauss’s theorem.
The differential form of Gauss’s law shows the relation between the electric field in space to the charge density ‘\[\rho \]’ at the point.
There are various limitations of Gauss’s law-
To imagine a Gaussian surface of a given shape is not explained by Gauss. According to Gauss, the Gaussian surface may be of any shape which should enclose any charge and this theorem can be applied to selected problems only.

Note: The area of the surface known as the vector direction is perpendicular to the surface. Electric flux is a scalar quantity. The electric flux through a closed surface is directly proportional to the electric charge that is when the charge increases then the flux also increases and vice versa.