Give Mathematical Derivation of electric flux =E. ds.

Content Curator

Electric flux is the measure of the flow of electric field through a surface. Mathematically, electric flux is defined as the dot product of electric field and area vector of the surface.

Consider a small area element dA on a surface with an electric field E. The electric flux dΦ through this area element is given by:

dΦ = E · dA

where · represents the dot product of E and dA. The electric field E is perpendicular to the area element dA, so the dot product of E and dA is simply the product of their magnitudes.

dΦ = E dA cos θ

where θ is the angle between the direction of the electric field and the normal to the surface. Since the electric field is perpendicular to the surface, the angle θ is 0 degrees and cos θ is 1. Therefore, we can simplify the equation to:

dΦ = E dA

Now, to find the total electric flux Φ through the entire surface, we integrate the electric flux over the entire surface area A:

Φ = ∫ E · dA

where the integral is taken over the entire surface. This is the mathematical expression of electric flux.

In summary, the electric flux Φ through a surface is given by the dot product of the electric field E and the area vector dA.

Electric Flux Derivation Example

Electric Flux Derivation Example

Also read:

Related Concepts
Gauss's Law	Charge transfer	Unit of Electric Field
Coulomb’s Law	Applications of Gauss’s Law	Continuous Charge Distribution
Force multiple charges	Electric Charges and Fields	Conservation of Charge

CBSE CLASS XII Related Questions

1.
Answer the following giving reason:
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensity.
(c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.

View Solution

2.
Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): In double slit experiment, if one slit is closed, diffraction pattern due to the other slit will appear on the screen.
Reason (R): For interference, at least two waves are required.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Both Assertion (A) and Reason (R) are false.

View Solution

3.
A coil of an AC generator, having 100 turns and area 0.1 m² each, rotates at half a rotation per second in a magnetic field of 0.02 T. The maximum emf generated in the coil is:

0.31 V
0.20 V
0.63 V
0.10 V

View Solution

4.
A proton moving with velocity \( V \) in a non-uniform magnetic field traces a path as shown in the figure. The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points P, Q, and R? What can you say about relative magnitude of magnetic fields at these points?

View Solution

5.
Prove that, in the Bohr model of the hydrogen atom, the time period of revolution of an electron in the \( n \)-th orbit is proportional to \( n^3 \).

View Solution

6.
A coil has 100 turns, each of area \( 0.05 \, \text{m}^2 \) and total resistance \( 1.5 \, \Omega \). It is inserted at an instant in a magnetic field of \( 90 \, \text{mT} \), with its axis parallel to the field. The charge induced in the coil at that instant is:

\( 3.0 \, \text{mC} \)
\( 0.30 \, \text{C} \)
\( 0.45 \, \text{C} \)
\( 1.5 \, \text{C} \)

View Solution

Give Mathematical Derivation of electric flux =E. ds.

CBSE CLASS XII Related Questions

3.
A coil of an AC generator, having 100 turns and area 0.1 m² each, rotates at half a rotation per second in a magnetic field of 0.02 T. The maximum emf generated in the coil is:

5.
Prove that, in the Bohr model of the hydrogen atom, the time period of revolution of an electron in the \( n \)-th orbit is proportional to \( n^3 \).

6.
A coil has 100 turns, each of area \( 0.05 \, \text{m}^2 \) and total resistance \( 1.5 \, \Omega \). It is inserted at an instant in a magnetic field of \( 90 \, \text{mT} \), with its axis parallel to the field. The charge induced in the coil at that instant is:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

Give Mathematical Derivation of electric flux =E. ds.

CBSE CLASS XII Related Questions

3.A coil of an AC generator, having 100 turns and area 0.1 m² each, rotates at half a rotation per second in a magnetic field of 0.02 T. The maximum emf generated in the coil is:

5.Prove that, in the Bohr model of the hydrogen atom, the time period of revolution of an electron in the \( n \)-th orbit is proportional to \( n^3 \).

6.A coil has 100 turns, each of area \( 0.05 \, \text{m}^2 \) and total resistance \( 1.5 \, \Omega \). It is inserted at an instant in a magnetic field of \( 90 \, \text{mT} \), with its axis parallel to the field. The charge induced in the coil at that instant is:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

3.
A coil of an AC generator, having 100 turns and area 0.1 m² each, rotates at half a rotation per second in a magnetic field of 0.02 T. The maximum emf generated in the coil is:

5.
Prove that, in the Bohr model of the hydrogen atom, the time period of revolution of an electron in the \( n \)-th orbit is proportional to \( n^3 \).

6.
A coil has 100 turns, each of area \( 0.05 \, \text{m}^2 \) and total resistance \( 1.5 \, \Omega \). It is inserted at an instant in a magnetic field of \( 90 \, \text{mT} \), with its axis parallel to the field. The charge induced in the coil at that instant is: