Electric Lines of Force-

Before you go through this article, make sure that you have gone through the previous article on Electric Lines of Force.

We have discussed-

• Electric lines of force represent the electric field around the charge.
• Electric lines of force are the imaginary curves.
• But the field they represent is real.

Electric Flux-

 The total number of electric lines of force passing perpendicularly through a given area held inside an electric field is called as electric flux.

Electric flux is denoted by the symbol ΦE.

The electric flux passing through any given area depends upon the following factors-

Also Read- Electric Lines of Force

1. Electric Field Strength-

Electric flux ∝ Strength of Electric Field

This is because-

• Stronger the electric field, closer are the electric lines of force.
• Closer the electric lines of force, greater the amount of electric flux passing through the surface.

2. Area of The Surface-

Electric flux ∝ Area of the Surface

This is because-

• Greater the area of the surface, greater the amount of electric lines of force passing through it.
• Therefor, greater is the electric flux passing through the area.

3. Angle between Electric Field & Area Vector-

Electric flux ∝ cosθ

This is because-

• As the angle (θ) between electric field and area vector increases, the number of electric lines of force passing through the area decreases.

On combining these factors, we get the following electric flux formula.

Also Read- Electric Field

Electric Flux Formula-

Electric flux ΦE passing through any given area $\overrightarrow{A}$ held inside an electric field of strength $\overrightarrow{E}$ is given by-

• The SI unit of electric flux is NC-1m2.
• Electric flux is a scalar quantity.

Electric Flux Density-

Electric flux passing per unit area through the surface is called as electric flux density.

Mathematically,

• The SI unit of electric flux density is NC-1.
• Electric flux density is a scalar quantity.

Important Notes-

It is worth mentioning the following notes-

Note-01:

• Area vector for a 2D surface is always perpendicular to its surface.
• Area vector for a 3D surface is always outward normal to its surface.

Note-02:

• The formula ΦE = $\overrightarrow{E}.\overrightarrow{A}$ = EAcosθ is valid if and only if the value of electric field E at every point lying on the area A is same.

Note-03:

• The formula ΦE = ∫$\overrightarrow{E}.\overrightarrow{dA}$ is valid everywhere.
• When electric field is uniform, $\overrightarrow{E}$ comes out of the integration and the formula reduces to ΦE = $\overrightarrow{E}.\overrightarrow{A}$.

Problem-01:

Calculate the electric flux passing through the surface kept in the electric field with its plane perpendicular to the electric field as shown-

Solution-

We have-

Given-

• Electric field strength (E) = 100 N/C
• Area of the surface (A) = 2 m x 2 m = 4 m2
• Angle between $\overrightarrow{E}$ and $\overrightarrow{A}$ = 0°

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

Substituting the values, we get-

ΦE = 100 x 4 x cos 0°

ΦE = 100 x 4 x 1

ΦE = 400 NC-1m2

Thus, electric flux passing through the given surface = 400 NC-1m2.

Problem-02:

Calculate the electric flux passing through the surface kept in the electric field with its plane parallel to the electric field as shown-

Solution-

We have-

Given-

• Electric field strength (E) = 100 N/C
• Area of the surface (A) = 2 m x 2 m = 4 m2
• Angle between $\overrightarrow{E}$ and $\overrightarrow{A}$ = 90°

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

Substituting the values, we get-

ΦE = 100 x 4 x cos 90°

ΦE = 100 x 4 x 0

ΦE = 0

Thus, no electric flux passes through the given surface.

Problem-03:

Calculate the electric flux passing through the circular disk of radius 7 m placed in the electric field as shown-

Solution-

We have-

Given-

• Electric field strength (E) = 100 N/C
• Radius of circular disk (R) = 7 m
• Angle between $\overrightarrow{E}$ and $\overrightarrow{A}$ = 60°

Now,

Area of the disk surface (A)

= πR2

= (22/7) x 7 x 7

= 154 m2

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

Substituting the values, we get-

ΦE = 100 x 154 x cos 60°

ΦE = 100 x 154 x 0.5

ΦE = 7700 NC-1m2

Thus, electric flux passing through the given circular disk = 7700 NC-1m2.

Problem-04:

If $\overrightarrow{E}$ = $6\hat{i}&space;+&space;3\hat{j}&space;+&space;4\hat{k}$, calculate the electric flux through a surface of area 20 units in Y-Z plane.

Solution-

Given-

• Electric field ($\overrightarrow{E}$) = $6\hat{i}&space;+&space;3\hat{j}&space;+&space;4\hat{k}$
• Area of the surface ($\overrightarrow{A}$) = $20\hat{i}$

We know, electric flux ΦE passing through any surface is given by-

ΦE = $\overrightarrow{E}.\overrightarrow{A}$

Substituting the values, we get-

ΦE = ($6\hat{i}&space;+&space;3\hat{j}&space;+&space;4\hat{k}$).$20\hat{i}$

ΦE = 120 units

Thus, electric flux passing through the given surface = 120 units.

Problem-05:

A cylinder of radius R is placed in a uniform electric field E with its axis parallel to the field as shown-

Calculate the electric flux passing through-

1. Left plane surface
2. Right plane surface
3. Curved surface
4. Complete cylinder

Solution-

Given-

• Electric field strength = E
• Radius of the cylinder = R

We know, electric flux ΦE passing through any surface is given by-

ΦE = EAcosθ

Part-01: Electric Flux Through Left Plane Surface-

Electric flux through left plane surface (Φ1)

= E x πR2 x cos 180°

= -EπR2

Thus, electric flux passing through the left plane surface = -EπR2.

Part-02: Electric Flux Through Right Plane Surface-

Electric flux through right plane surface (Φ2)

= E x πR2 x cos 0°

= EπR2

Thus, electric flux passing through the right plane surface = EπR2.

Part-03: Electric Flux Through Curved Surface-

Electric flux through curved surface (Φ3)

= E x 2πRL x cos 90°

= 0

Thus, electric flux passing through the curved surface = 0.

Part-04: Electric Flux Through Complete Cylinder-

Electric flux through complete cylinder (ΦE)

= Φ1 + Φ2 + Φ3

= -EπR2 + EπR2 + 0

= 0

Thus, electric flux passing through the complete cylinder = 0.

Problem-06:

A positive charge of 17.7 μC is placed at the center of a hollow sphere of radius 0.5 m. Calculate the flux density through the surface of the sphere.

Solution-

Given-

• Radius of hollow sphere (R) = 0.5 m
• Charge at center (Q) = 17.7 μC

Step-01: Calculating Electric Field At Surface-

Electric field at the surface of a hollow sphere due to the charge kept at the center is constant.

The electric field at the surface is directed outward normal.

It is given by-

Substituting the values, we get-

Electric field at surface

= (9 x 109 x 17.7 x 10-6) / (0.5)2

= 637.2 x 103

= 6.37 x 105 N/C

Step-02: Calculating Electric Flux Through Surface-

We have-

Electric flux passing through surface of the sphere

= EAcosθ

= E x 4πR2 x cosθ

= 6.37 x 105 x (4 x 3.14 x 0.5 x 0.5) x cos0°

= 20 x 105 NC-1m2

Step-03: Calculating Electric Flux Density-

Electric flux density

= Total electric flux passing through the surface / Area of the surface

= 20 x 105 / (4 x 3.14 x 0.5 x 0.5)

= 20 x 105 / 3.14

= 6.36 x 105

= 6.4 x 105 N/C

Thus, electric flux density through the surface = 6.4 x 105 N/C.

To gain better understanding about Electric Flux,

Watch this Video Lecture

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