Electrostatic | Physics and Universe https://physicsanduniverse.com Sat, 28 Feb 2015 05:40:36 +0000 en-US hourly 1 https://wordpress.org/?v=6.8.6 52870843 Gauss Theorem https://physicsanduniverse.com/gauss-theorem/ https://physicsanduniverse.com/gauss-theorem/#respond Sat, 28 Feb 2015 05:40:36 +0000 http://physicsanduniverse.com/?p=7879 Gauss theorem (German mathematician Carl Friedrich Gauss) is used to relate flux (the number of lines of force crossing an area normally is called electric flux or simply flux) through a closed surface to the charge enclosed by it. It states that The total electric flux through an arbitrary closed surface, enclosing the charge is equal to the ratio charge to the permittivity of the medium.

Let us consider a point charge +Q inside a closed surface and let dA be elemental area on the closed surface at a distance of r from the point charge +Q. Then the electric field intensity at the area dA is gauss theorem

E = \dfrac{Q}{4\pi \epsilon r^2}

Let is assume that the direction of E subtends an angle \theta with outward normal to dA. Then the component of E normal to dA is

E_{normal} = Ecos\theta

Now the flux through dA is

d\phi =E_{normal} \times dA = E cos\theta dA

Putting the value of E

d \phi = \dfrac{Q}{4\pi\epsilon {r^2}}{dAcos\theta}

d \phi = \dfrac{Q}{4\pi\epsilon}{\dfrac{dAcos\theta}{r^2}}

The factor \dfrac{dAcos\theta}{r^2} is called solid angle subtended by dA at +Q and denoted by d\omega . Hence

d\omega = \dfrac{Q}{4\pi \epsilon} \dfrac{dAcos\theta}{r^2}

The total flux through the closed surface is obtained by integrating the equation over the complete solid angle

\int d\phi = \phi = \dfrac{Q}{4 \pi \epsilon} \int^{4\pi}_0{d\omega}

It should be noted that the minimum value of solid angle is zero and maximum value is 4\pi . So,

\phi =\dfrac{Q}{4\pi\epsilon} 4\pi = \dfrac{Q}{\epsilon}

this proves the gauss’s theorem of any arbitrary closed surface.

It should be noted that the Gauss theorem shows that the total flux crossing any closed surface drawn around a point charge is a constant and is independent of shape and size of the surface.

 

]]> https://physicsanduniverse.com/gauss-theorem/feed/ 0 7879 Coulomb’s Law https://physicsanduniverse.com/coulombs-law/ https://physicsanduniverse.com/coulombs-law/#respond Sun, 08 Feb 2015 09:39:21 +0000 http://physicsanduniverse.com/?p=7872 We know that charged bodies exerts force on each other and force between these two charges is known as the electric interaction. An experimentation and studies of these charges have shown that

  1. There are only two types of charges, positive and negative
  2. Unlike charges attract each other
  3. Like charges repel each other

Coulomb studied about these charges and his conclusion is that if the charged bodies were small as compared to distance between them, then the force between them depends on the magnitude of charge and distance between them. After experimentation, he formulated a law determining the force between charges called coulomb’s law. According to this law the force between two charges is

  1. directly proportional to the product of the quantity of charges
  2. inversely proportional to the square of the distance between them
  3. the direction of force lies along the line joining the centers of charged body

Let Q_1 and Q_2 be magnitude of charges which are separated by the distance r then we have

F \propto Q_1 Q_2

F \propto \dfrac{1}{r^2}

combining above two equations we get

F \propto \dfrac{Q_1Q_2}{r^2}

F = k\dfrac{Q_1Q_2}{r^2}

where k is proportionality constant called permittivity.

Permittivity

The permittivity k depends on the system of unit used and medium between charges. In SI unit the charge is measured in coulomb, force in Newton and distance in meter. So, permittivity (k) is given by

k = \dfrac{1}{4\pi\epsilon_0}

where \epsilon_0 is a constant called permittivity of free space. Using this value of k in force equation we got above, we have

F = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q_1Q_2}{r^2}

From this we can see that the SI unit of \epsilon_0 is (coulomb)^2(Newton)^{-1}(meter)^{-2} . The value of \epsilon_0 is 8.85 \times 10^{-12}Fm^{-1} . So,

\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9

It should be noted that permittivity of the medium differs. If charges are for example placed in water or oil, the force between charges is reduced. So, for a medium k is written as

k = \dfrac{1}{4\pi\epsilon}

Here \epsilon is called permittivity of the medium. Its value is different for different medium.

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