ELECTRIC FLUX

ELECTRIC FLUX:-

                       Number of electric lines of forces passing through a certain area is called electric flux. Greater the electric field lines passing greater is electric flux and is represented by Φ_{E  and is given by relation}
    Φ_{E =} E A cos ------------(i)

= Electric field intensity                                                                                                                                            

= Vector area


NOTE:-
For any cross sectional area if a normal is drawn on it then its is called vector area.

 SI unit of electric flux is Nm²/C

Or  Kg m³ s^-3 A^-1

Dimensions of electric flux is [ML³T^-3A^-1]

TYPES:-

• MAXIMUM & POSITIVE ELECTRIC FLUX:-

 If the surface area is held perpendicular to electric field lines such that vector area is parallel to these  lines

 that is  θ=0 degrees between&as shown in fig then the magnetic flux  is maximum and positive  

put  θ=0 in (i) we get

       Φ_{E =} E A cos 0

  Φ_{E =} E A-----(ii)

(ii) gives the relation of maximum electric flux.

• MINIMUM & NEGATIVE ELECTRIC FLUX:-

          If the surface area is held perpendicular to electric field lines so that its vector area is anti parallel to these lines that is θ= 180 degrees between& as shown in the fig then electric flux is negative and minimum.

  put  θ= 180 in (i) we get 

  Φ_{E =} E A cos 180

Φ_E= -EA-----(iii)

(iii) shows the relation of minimum and negative electric flux.

• ZERO ELECTRIC FLUX:-

              If surface area is held parallel to electric field lines so that vector area is perpendicular to electric field lines that is θ=90 degrees between & as shown in the fig then electric flux is zero

put  θ=90 in (i) we get  

 Φ_{E =} E A cos 90

Φ_E=0----(iv)

_{(iv) represents equation for zero electric flux}

• FOR ANY ANGLE:-

If surface area is held to electric field lines so that its vector area is making some angle θ to these lines then

electric flux can have any value. If surface area is rotated from 0 to 90 degrees electric flux increases and if vector area is rotated from 90 to 0 degrees electric flux increases  and vice versa

_{now (i) can} 

 Φ_E=A_x.E

 Φ_{E= A} cos   E

  Φ_{E =} E A cos -----(v)

PROBLEMS:-

NUMBER I

A flat surface of area 3.20m² is rotated in a uniform electric field of magnitude E = 6.20 x 10⁵ N/C. Determine the electric flux through this (a) when the electric filed is perpendicular to the surface and (b) when the electric field is parallel to the surface

SOLUTION:-

(a) When the electric filed is perpendicular to the surface 
Angle between the vector E and vector area A is zero and cos 0 = 1 and hence Flux is  6.20 x 10⁵x3.2
= 1.98410⁶ (Nm²/C) 

b) When the electric field is parallel to the surface. 
Angle between the vector E and vector area A is 90 and cos 90 =0 and hence the Flux is zero. .