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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CBSE CLASS XII Related Questions

  • 1.
    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:

      • \( 0.5 \times 10^3 \, \text{m}^{-1} \)
      • \( 6.0 \times 10^2 \, \text{m}^{-1} \)
      • \( 7.5 \times 10^2 \, \text{m}^{-1} \)
      • \( 1.5 \times 10^3 \, \text{m}^{-1} \)

    • 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

        • Production of AC is economical.
        • AC can be easily and efficiently converted from one voltage to another.
        • AC can be transmitted economically over long distances.
        • AC is less dangerous.

      • 3.
        Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

          • 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.

        • 4.
          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:

            • 5 V
            • 3.5 V
            • 2.5 V
            • Zero V

          • 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.

              • 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.

            • 6.
              A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
              the magnetic dipole moment of the coil, and
              the magnitude of the counter torque that must be applied to prevent the coil from turning.

                CBSE CLASS XII Previous Year Papers

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