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Electric Flux

From: Nancy James

Subject: Electric flux
Message Body:
Could some one please explain what is written in the ncert book of PHYSICs class XII about ‘electric flux’?

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What are the similarities and differences in magnetic and electric field?

Similarities and differences in magnetic and electric field (Dilpreet posted this question)

Similarities Differences
 Both electric and magnetic field are conservative forcesBoth obey inverse square law

Both are non contact forces (Forces can be exerted without contact)

Both are attractive as well as repulsive (Like poles repel, like charges repel; unlike poles attract, unlike charges attract)

 Electric field is produced by a charge whether at rest or in motionBut magnetic field is produced only by a moving charge

 

The total magnetic flux through any closed surface is always zero, but the total electric flux through any closed surface is equal to the net charge enclosed by the surface multiplied by the reciprocal of absolute permitivity

Electric field lines are discontinuous as they have a starting point (+ charge) and an ending point (- charge); But magnetic field lines are continuous, they always form closed loops

 

Radius of an electron and Distance between electric field lines.

“What is the distance between two lines of force produced by a charge or the distance is zero between them and what is the diameter of electron?” – Ela posted.

Answer:

The electric field lines are imaginary lines used to represent the electric field visually. The electric field lines are imagined or drawn relatively closer where there is greater electric field. (Though electric flux is often defined as the number of field lines per unit area, it is not so. Electric flux cannot be talked in terms of just numbers. That definition is just to make  a quick idea related to the lines in a given situation. To cross check, just calculate the electric flux from a proton and try to draw that much number of lines around a point)

The radius of electron has been calculated from Mass Energy relation as 2.817 940 3267 x 10-15m

Therefore, Diameter of an electron = 5.635880578916 x 10-15m

Please note that the above value is just a theoretical estimation. The actual size of electron is not experimentally determined.

Refer to http://en.wikipedia.org/wiki/Classical_electron_radius for more

Gauss Theorem – Electrostatics

DIVYA asked:

“WHAT IS THE  MAGNITUDE OF ELECTRIC  FIELD IN THE  GAUSSIAN SURFACE OF  A CUBE,AT ITS FACE ,AT ITS CENTRE, AT ITS CORNERS OR AT ANY OTHER POINT INSIDE THE CUBE. ALSO TELL ME THE THE WAY TO KNOW IT.”

Ans:

According to Gauss Theorem, the total electric flux through a closed surface,\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A}  = \frac{Q}{\varepsilon_0},

The Gaussian surface is an imaginary surface. So, for calculating the electric field at a point using Gauss theorem, we have to imagine a Gaussian surface symmetric with the given charge distribution.

I have assumed that there is a point charge Q at the centre of the cube.

At its face (at a point on the face on the line radially outwards from the point charge at the centre of the cube and perpendicular to the face)

The distance is equal to half the length of side of the cube (Say L).

Therefore the electric field,image  

 

On the corner,

Calculate the distance from the centre of the cube to its corner and replace (L/2) in the above equation with that distance.

A Question from Electric Flux and Gauss Theorem

Zeenath asked:

How can you prove that dS in φ =E.dS is dS Cos θ if the area is tilted at an angle ? I need the mathematical Steps.

 

Ans;

Dear Zeenath,
Electric Flux is defined as the total no of field lines passing normal to the surface. So while calculating, we need to consider the area of the surface normal to the field lines only. That is why we take the dot product of E and ds, where ds is the area vector (not just the area: Area vector is a vector whose magnitude is equal to the area and dirtected normal to the surface.)
Then by definition of dot product,
dφ =E.dS = EdS cos θ
which gives the component of E and the component of area vector in the direction of E (When area vector is in the direction of E, the actual component of area is perpendicular to E)
Hope the matter is clear.
Mathew Abraham

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