Give Mathematical Derivation of electric flux =E. ds.

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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

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Give Mathematical Derivation of electric flux =E. ds.

CBSE CLASS XII Related Questions

1.
Light of which of the following colours will have the maximum energy in a photon associated with it?

2.
The electric potential (V ) and electric field (⃗ E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the

4.
The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

5.
Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

6.
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Give Mathematical Derivation of electric flux =E. ds.

CBSE CLASS XII Related Questions

1.Light of which of the following colours will have the maximum energy in a photon associated with it?

2.The electric potential (V ) and electric field (⃗ E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the

4.The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

5.Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

6.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Light of which of the following colours will have the maximum energy in a photon associated with it?

2.
The electric potential (V ) and electric field (⃗ E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the

4.
The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

5.
Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

6.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be: