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Electric Flux

Learn concept of electric flux definition with examples and equation. Practice electric flux formula and calculate electric flux through a cylinder.

Concept of Flux

  • Flux of any physical quantity is the measure of flow of that quantity passing through any specific area.
  • To understand the idea of flux, consider a region in which rain is falling vertically downwards and ring is placed in this region.
  • The rate of flow of rain passing through area of the ring depends on its area and orientation of ring.

Case I : "Rings of Different Area and Same Orientation"

Since, 

Area c > Area b > Area a

Hence, flux through

c > b > a

Case II : "Rings of Same Area but Different Orientation"

  • Plane of ring (a) oriented perpendicular to field.
  • Plane of ring (b) oriented at an angle to the normal of the field.
  • Plane of ring (c) oriented along the field.
  • Flux through Ring (a) > Ring (b) > Ring (c)
  • Flux of Ring (c) is zero as its plane is oriented along the field so, no rain is passing through its area.

Electric Flux

Electric flux is defined as the number of field lines passing through an area.

Illustration Questions

Three identical rings of area A are placed in an uniform electric field, as shown in figure. Choose the correct option regarding to flux linked with them.

A c > b > a

B a = b = c

C a > b > c

D a > c > b

×

Area of rings is same, but orientation is different.

image

Number of field lines passing through (a) > Number of field lines passing through (b) > Number of field lines passing through (c)

image

Hence, flux linkage is in order

Ring (a) > Ring (b) > Ring (c)

image

Three identical rings of area A are placed in an uniform electric field, as shown in figure. Choose the correct option regarding to flux linked with them.

image
A

c > b > a

.

B

a = b = c

C

a > b > c

D

a > c > b

Option C is Correct

Flux Linked with an Area

  • Consider a closed surface of cylinder placed in a non-uniform electric field as shown in figure.
  • Area of P = Area of Q
  • Number of field lines associated with P is greater than number of field lines associated with Q.
  • Flux linked with P is greater than flux linked with Q.
  • Flux is denoted by \(\phi\).
  • Hence, \(\phi_P>\phi_Q\)   

Entering and Leaving Concept of Field Lines for Calculation of Flux

  • Consider a body placed in an electric field.
  • Calculate number of field lines entering and leaving the body.
  • Case 1: If number of field lines entering and leaving the body are equal,

           then total flux linked with the body will be zero.

  • Case 2: If number of field lines entering and leaving are not equal,

          then, Net flux= number of field lines leaving - number of field lines entering

Concept of Projected Area

  • When any body is placed in a uniform electric field such that its surface is perpendicular to the electric field, then electric flux associated with it, is maximum.
  • Now if, surface of same body is placed at some angle with the electric field rather than perpendicular, then electric flux associated with it, will be less than maximum value.
  • The number of electric field lines passing through the tilted surface is less than the number of field lines passing through perpendicular surface but same as the number of field lines passing through the projection of area of surface.
  • Number of field lines passing through tilted surface (A) = Number of field lines passing through projection of area of surface.

    \(\phi_{E_\perp'}=\phi_{E'}<\phi_E\)

    \(A_{E_\perp'} < A_{E'} = A_E\)    

Illustration Questions

Choose the correct option regarding flux linkage among surface P and surface Q of the cylinder placed in a non-uniform electric field, as shown in figure.

A \(\phi_P>\phi_Q\)

B \(\phi_Q>\phi_P\)

C \(\phi_P=\phi_Q\)

D

×

Number of field lines passing through (P)  > Number of field lines passing through (Q)

Hence, flux linkage \(\phi_P>\phi_Q\)

Choose the correct option regarding flux linkage among surface P and surface Q of the cylinder placed in a non-uniform electric field, as shown in figure.

image
A

\(\phi_P>\phi_Q\)

.

B

\(\phi_Q>\phi_P\)

C

\(\phi_P=\phi_Q\)

D

Option A is Correct

Mathematical form of Flux

Area Vector

By convention, area vector is always taken outwards / perpendicular to the plane as shown in figure.

  • Consider a closed surface, as shown in figure. Area vector for all the three surface A, B and C of cylinder is taken outwards/perpendicular to the surface.

Flux

Consider a ring placed in a uniform electric field, as shown in figure.

  • Electric flux linked with the area A is given as the dot product of electric field and area vector.
  • Flux linked with area A   

                                             \(\phi_E=E(A\;cos\theta)\)

                                                  \(\phi_E=\vec E\cdot\ \vec A\)   

  • For non-uniform electric field, electric field may vary over a large surface.
  • The concept of flux is meaningful for small area where electric field is approximately constant, when total surface is placed in non-uniform field.

\(\Delta\phi_E=E(\Delta A\;cos\theta)\)

or \(\Delta\phi_E=\vec E\cdot\Delta \vec A\)

where, \(\Delta\phi_E\) is the electric field through this element.

  • The electric flux through whole surface is given as summation of electric flux through all these small elements.

\(\phi_E=\sum\vec E\cdot\Delta\vec A\)

  • If area of each element approaches zero, the number of elements approach infinity, then the sum is replaced by integral.

\(\phi_E=\int_s\vec E\cdot d\vec A\)

  • For closed surface \(\phi_E=\int\vec E\cdot d\vec A\)
  • If in a region electric field is given as \(\vec E=E_x\hat i+E_y\hat j+E_z\hat k\)
  • An area A is placed in an electric field with area vector \(\vec A=A_x\hat i+A_y\hat j+A_z\hat k\)
  • Flux linked with this area is given as \(\phi=\vec E\cdot\vec A\)
  • \(\phi=E_xA_x +E_yA_y +E_zA_z\)      

Illustration Questions

Calculate the amount of flux linked with area vector \(\vec A = (3\hat i+4\hat j)\,m^2\)  which is placed in an electric field \(\vec E = (4\hat i+6\hat j)\,N/C\).

A 0 Nm2/C

B 36 Nm2/C

C 30 Nm2/C

D 400 Nm2/C

×

Flux linked with area A is given as \(\phi=\vec E\cdot\vec A\)

Given: \(\vec E = (4\hat i+6\hat j)\,N/C\) , \(\vec A = (3\hat i+4\hat j)\,m^2\) 

\(\phi=(4\hat i+6\hat j)\cdot(3\hat i+4\hat j)\)

\(\phi=\)12 + 24 = 36 Nm2/C

Calculate the amount of flux linked with area vector \(\vec A = (3\hat i+4\hat j)\,m^2\)  which is placed in an electric field \(\vec E = (4\hat i+6\hat j)\,N/C\).

A

0 Nm2/C

.

B

36 Nm2/C

C

30 Nm2/C

D

400 Nm2/C

Option B is Correct

Calculation of Electric Flux when Angle is Given

Area Vector

By convention, area vector is always taken outwards / perpendicular to the plane as shown in figure.

  • Consider a closed surface, as shown in figure. Area vector for all the three surface A, B and C of cylinder is taken outwards/perpendicular to the surface.

Flux

Consider a ring placed in a uniform electric field, as shown in figure.

  • Electric flux linked with the area A is given as the dot product of electric field and area vector.
  • Flux linked with area A 

\(\phi_E=E(A\;cos\theta)\)

\(\phi_E=\vec E\cdot\vec A\)     

Illustration Questions

Calculate electric flux through the surface given E = 600 N/C, A = 10 cm×10 cm and \(\theta=\)60° as shown in figure.

A 300 Nm2/C

B 3 Nm2/C

C 3000 Nm2/C

D 0

×

Electric flux through a given surface A is given as  \(\phi=E(A\;cos\theta)\)

Given:

E = 600 N/C,

A = 10 ×10 cm2 = 10–2 m2

 \(\theta=\)60°

\(\phi=600\times10^{-2}cos60°\)

\(\phi=6\times\dfrac {1}{2}=3\) Nm2/C

Calculate electric flux through the surface given E = 600 N/C, A = 10 cm×10 cm and \(\theta=\)60° as shown in figure.

image
A

300 Nm2/C

.

B

3 Nm2/C

C

3000 Nm2/C

D

0

Option B is Correct

Flux Linked with Cylindrical Body

  • Consider a cylindrical body placed in a uniform electric field \(\vec E\).
  • To calculate flux, by convention area vector is always taken perpendicular and in outward direction to the surface.

  • Area vector of three surface of a cylindrical body, are shown in figure.

Flux through A1

By convention area vector is always taken perpendicular and in outward direction to the surface.

Angle between \(\vec E\) and \(d\vec A_1\) is 180°.

\(\phi_1=\vec E\cdot d\vec A_1\)

\(\phi_1=E\;dA_1\;cos(180°)\)

\(\phi_1=-E\, dA_1\)

  • It can be concluded that for surface 1 by convention, we have taken the \(d\vec A\) outside the surface which is in opposite direction to electric field.
  • Hence, the flux coming out of surface 1 is negative.

 

CONCLUSION: For any particular surface through which the electric field lines are entering, flux will be negative.

Flux through A2

Angle between \(\vec E\) and \(d\vec A_2\) is 90°.

\(\phi_2=\vec E\cdot d\vec A_2\)

or, \(\phi_2=E\;dA_2\;cos(90°)\)

or, \(\phi_2=0\)   

CONCLUSION: If electric field and area vector are perpendicular to each other, then flux through that surface is zero.

Flux through A3

Angle between \(\vec E\) and \(d\vec A\) is 0°.

\(\phi_1=\vec E\cdot d\vec A\)

\(\phi_1=E\;dA\;cos0°\)

\(\phi_1=E\, dA\)   

 

  • By convention, for surface 3 we have taken \(d\vec A\) outside the surface which is in the direction of electric field.
  • Hence, flux coming out of surface 3 is positive.

CONCLUSION: For any particular surface through which the electric fields are leaving, flux will be positive.

Illustration Questions

A cube is placed in a uniform electric field \(\vec E=E_x\hat i+E_z\hat k\) N/C. Determine the sign of flux through ABCD, CDHG and EFGH.

A –ve, –ve, –ve 

B +ve, +ve, –ve 

C +ve, +ve, +ve

D –ve, –ve, +ve

×

Electric field is present in +X and +Z direction.

For face ABCD

Direction of area vector

\(\hat n_1=\hat k\)

\(\vec A_1=A_1\;\hat k\;m^2\)

\(\vec E=(E_x\hat i + E_z\;\hat k)\) N/C

Total flux  

\(\phi_1=(E_x\hat i +E_z \hat k)\cdot(A_1\hat k)\)

\(\phi_1=E_zA_1=\) positive

image

For face CDHG

Direction of area vector

\(\hat n_2=\hat i\)

\(\vec A_2=(A_2\;\hat i)\;m^2\)

\(\vec E=(E_x\hat i + E_z\;\hat k)\) N/C

Total flux  

\(\phi_2=(E_x\hat i +E_z \hat k)\cdot(A_2\hat i)\)

\(\phi_2=E_xA_2=\) positive

image

For face EFGH

Direction of area vector

\(\hat n_3=-\hat k\)

\(\vec A_3=(-A_3\;\hat k)\;m^2\)

\(\vec E=(E_x\;\hat i + E_z\;\hat k)\) N/C

Total flux  

\(\phi_3=(E_x\hat i +E_z \hat k)\cdot(-A_3\hat k)\)

\(\phi_3=-E_z\;A_3=\) negative

image

A cube is placed in a uniform electric field \(\vec E=E_x\hat i+E_z\hat k\) N/C. Determine the sign of flux through ABCD, CDHG and EFGH.

image
A

–ve, –ve, –ve 

.

B

+ve, +ve, –ve 

C

+ve, +ve, +ve

D

–ve, –ve, +ve

Option B is Correct

Concept of Projected Area

  • When any body is placed in a uniform electric field such that its surface is perpendicular to the electric field, then electric flux associated with it, is maximum.
  • Now if, surface of same body is placed at some angle with the electric field rather than perpendicular, then electric flux associated with it, will be less than maximum value.
  • The number of electric field lines passing through the tilted surface is less than the number of field lines passing through perpendicular surface but same as the number of field lines passing through the projection of area of surface.
  • Number of field lines passing through tilted surface (A) = Number of field lines passing through projection of area of surface
  •  

    \(\phi_{E'_\perp}=\phi_{E'}<\phi_E\)

    \(A_{E'_\perp} < A_{E'} = A_E\)     

    \(\phi=E(A\;cos\theta)\)

    or, \(\phi=\) (Electric field) (projection of area perpendicular to electric field)

Illustration Questions

The prism of side 'a' is placed in uniform electric field \(\vec E=\vec E_0\;\hat k\) N/C, find the flux linked with surface EFBC, if unit area vector of \(\vec A_{EFBC}\) is \(\dfrac {\hat j-\hat k}{\sqrt2}\).

A \(E_0a^2\)

B \(E_0a^2\dfrac {\sqrt3}{2}\)

C \(E_0a\)

D \(E_0/a\)

×

Area = \(\sqrt{2}a(a)=\sqrt{2}a^2\) (from co-ordinates)

\(\vec A=\dfrac {\sqrt2a^2(\hat j-\hat k)}{\sqrt 2}\)

= \(a^2(\hat j-\hat k)\)

image

\(\phi=\vec E\cdot\vec A\)

\(=E_0\;\hat k\cdot a^2(\hat j-\hat k)\)

\(=-E_0\; a^2\)(entering)

image

Magnitude of flux = \(E_0\;a^2\)

image

The prism of side 'a' is placed in uniform electric field \(\vec E=\vec E_0\;\hat k\) N/C, find the flux linked with surface EFBC, if unit area vector of \(\vec A_{EFBC}\) is \(\dfrac {\hat j-\hat k}{\sqrt2}\).

image
A

\(E_0a^2\)

.

B

\(E_0a^2\dfrac {\sqrt3}{2}\)

C

\(E_0a\)

D

\(E_0/a\)

Option A is Correct

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