General Physics> explain what is gauss law? and give the e...

explain what is gauss law? and give the examples ? and what is the applycations of the gauss law?

suresh , 8 Years ago

Grade 11

3 Answers

raju

Last Activity: 8 Years ago

Outline of proof The integral form of Gauss's law is: $= \frac{Q}{\varepsilon_0}$ $\oiint$ ${\scriptstyle S}$ $\mathbf{E} \cdot \mathrm{d}\mathbf{A}$

for any closed surface S containing charge Q. By the divergence theorem, this equation is equivalent to:

$\iiint\limits_V \nabla \cdot \mathbf{E} \ \mathrm{d}V = \frac{Q}{\varepsilon_0}$

for any volume V containing charge Q. By the relation between charge and charge density, this equation is equivalent to:

$\iiint\limits_V \nabla \cdot \mathbf{E} \ \mathrm{d}V = \iiint\limits_V \frac{\rho}{\varepsilon_0} \ \mathrm{d}V$

for any volume V. In order for this equation to be simultaneously true for every possible volume V, it is necessary (and sufficient) for the integrands to be equal everywhere. Therefore, this equation is equivalent to:

$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}.$

Thus the integral and differential forms are equivalent.

Approved

raju

Last Activity: 8 Years ago

The electric charge that arises in the simplest textbook situations would be classified as "free charge"—for example, the charge which is transferred in static electricity, or the charge on a capacitor plate. In contrast, "bound charge" arises only in the context of dielectric (polarizable) materials. (All materials are polarizable to some extent.) When such materials are placed in an external electric field, the electrons remain bound to their respective atoms, but shift a microscopic distance in response to the field, so that they're more on one side of the atom than the other. All these microscopic displacements add up to give a macroscopic net charge distribution, and this constitutes the "bound charge".

Although microscopically, all charge is fundamentally the same, there are often practical reasons for wanting to treat bound charge differently from free charge. The result is that the more "fundamental" Gauss's law, in terms of E (above), is sometimes put into the equivalent form below, which is in terms of D and the free charge only.

Integral form

This formulation of Gauss's law states the total charge form:

$\Phi_D = Q_\text{free}\!$

where Φ_D is the D-field flux through a surface S which encloses a volume V, and Q_free is the free charge contained in V. The flux Φ_D is defined analogously to the flux Φ_E of the electric field E through S:

$\Phi_{D} =$ $\oiint$ ${\scriptstyle S}$ $\mathbf{D} \cdot \mathrm{d}\mathbf{A}$

Differential form

The differential form of Gauss's law, involving free charge only, states:

$\mathbf{\nabla} \cdot \mathbf{D} = \rho_\text{free}$

where ∇ · D is the divergence of the electric displacement field, and ρ_free is the free electric charge density.

raju

Last Activity: 8 Years ago

definition:-

^[3].The net electric flux through any closed surface is equal to ¹⁄_ε times the net electric charge enclosed within that closed surface

Gauss's law has a close mathematical similarity with a number of laws in other areas of physics, such as Gauss's law for magnetism and Gauss's law for gravity. In fact, any "inverse-square law" can be formulated in a way similar to Gauss's law: For example, Gauss's law itself is essentially equivalent to the inverse-square Coulomb's law, and Gauss's law for gravity is essentially equivalent to the inverse-square Newton's law of gravity.

Gauss's law is something of an electrical analogue of Ampère's law, which deals with magnetism.

The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss's theorem. Each of these forms in turn can also be expressed two ways: In terms of a relation between the electric field E and the total electric charge, or in terms of the electric displacement field D and the free electric charge.^[4]