Electric Flux and Dipole

Electric Flux:

Electric Flux is defined as a number of electric field lines, passing per unit area.

Consider a plane surface of the area ?S in a uniform electric field E. A positive normal N to the surface makes angle $\theta$ with the electric field E.

Figure:4.a

Electric flux of the electric field through ?S is given by,$\Delta\phi =E\Delta S cos \theta $

From the above equation, an area vector ΔS of magnitude ΔS can be introduced along the positive normal. Then, $\Delta \phi =E. \Delta S$

The direction of area vector is always along normal to the surface.

If $\theta =90,$then $\Delta \phi =E\Delta S cos 90=0$

The electric flux becomes zero if the normal to the surface is perpendicular to the electric field.

Figure:4.b

Since the strength of electric field is directly proportional to a number of lines passing per unit area, electric flux also signifies for determining the strength of electric field. The S.I unit of electric flux is $Nm^{2}C^{1}$.

Electric dipole:

Electric Dipole is a couple of opposite charges $q$ and $-q$, separated by a distance, say d. The line joining the two charges gives the direction of the electric dipole in space. The direction of the dipole is from negative charge $-q$ to positive charge $q$. The unit of dipole moment is coulomb-meter (Cm).

Figure:4.c

The electric dipole is denoted by ${p}$ and equals to the product of the magnitude of charge and distance between the $q$ and $-q$.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>p</mi><mo>→</mo></mover><mo>=</mo><mi>q</mi><mo> </mo><mover><mi>d</mi><mo>→</mo></mover></math>

Electric field due to dipole:

Figure:4.d

The electric field due to dipole is called dipole field. The net electric field due to dipole will be the vector sum of the electric field due to charge and .

Electric Field along the axis of the dipole:

 

Electric Field along the axis of the dipole

Figure:4.e

Let P be a point at a distance $r$ from the centre of a dipole.

Then electric field due to charge $q$ is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>E</mi><mn>1</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mi>q</mi><msup><mfenced><mrow><mi>r</mi><mo>-</mo><mi>a</mi></mrow></mfenced><mn>2</mn></msup></mfrac><mover><mi>p</mi><mo>^</mo></mover></math>

Electric field due to charge is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>E</mi><mn>2</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><mo>-</mo><mi>q</mi></mrow><msup><mfenced><mrow><mi>r</mi><mo>+</mo><mi>a</mi></mrow></mfenced><mn>2</mn></msup></mfrac><mover><mi>p</mi><mo>^</mo></mover></math>

The resultant electric field is given by,

Electric field on the equatorial plane:

Figure:4.f

From the above figure, Then the magnitude of electric field at P due to charge $q$ and $-q$ is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>E</mi><mn>1</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mi>q</mi><mfenced><mrow><msup><mi>r</mi><mn>2</mn></msup><mo>+</mo><msup><mi>a</mi><mn>2</mn></msup></mrow></mfenced></mfrac><mover><mi>p</mi><mo>^</mo></mover></math>

and

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>E</mi><mn>2</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mi>q</mi><mfenced><mrow><msup><mi>r</mi><mn>2</mn></msup><mo>+</mo><msup><mi>a</mi><mn>2</mn></msup></mrow></mfenced></mfrac><mover><mi>p</mi><mo>^</mo></mover></math>

From the diagram, it is clear that the vertical component of electric field $E_{1}sin \theta $and $E_{2}sin \theta $will cancel out. Only the horizontal component of field will act and hence, the resultant electric field is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>E</mi><mo>=</mo><mo>-</mo><mfenced><mrow><msub><mi>E</mi><mn>1</mn></msub><mo>+</mo><msub><mi>E</mi><mn>2</mn></msub></mrow></mfenced><mi>cos</mi><mo> </mo><mi>θ</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>E</mi><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><mn>2</mn><mi>q</mi><mi>a</mi></mrow><mfenced><mrow><msup><mi>r</mi><mn>2</mn></msup><mo>+</mo><msup><mi>a</mi><mn>2</mn></msup></mrow></mfenced></mfrac><mover><mi>p</mi><mo>^</mo></mover><mspace linebreak="newline"/><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>=</mo><mo> </mo><mo>-</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><mn>2</mn><mi>q</mi><mi>a</mi></mrow><msup><mi>r</mi><mn>3</mn></msup></mfrac><mover><mi>p</mi><mo>^</mo></mover></math>

(as $r>>a$)

The above equation shows that dipole depends on the product of $qa$. Hence, the dipole moment vector $p$ of an electric dipole is defined by $p=q\times 2q {p}$.

 

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