An electric dipole of dipole moment 20 × 10 - 6 C is enclosed by a closed surface. What is the net flux coming out of the surface?

According to question , P → = 20 × 10 - 6 C From Gaussian law, flux is given by, ϕ = q i n ε o And we know that a dipole have both −ve and +ve charge of equal magnitude. Which implies that the total enclosed charge is given by, q i n = 0 ⇒ ϕ = 0 ε o = 0 &there4; There will be no flux through the surface.

An electric dipole of dipole moment 20 × 10 - 6 C is enclosed by a closed surface. What is the net flux coming out of the surface?

According to question , P → = 20 × 10 - 6 C From Gaussian law, flux is given by, ϕ = q i n ε o And we know that a dipole have both −ve and +ve charge of equal magnitude. Which implies that the total enclosed charge is given by, q i n = 0 ⇒ ϕ = 0 ε o = 0 &there4; There will be no flux through the surface.

E M B I B E

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 4

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An electric dipole of dipole moment $20 \times 10^{- 6} C$ is enclosed by a closed surface. What is the net flux coming out of the surface?

Important Questions on Gauss' Law and its Application

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12th West Bengal Board

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 5

Number of field lines entering a Gaussian surface is less than the number of field lines emerging from it. What it implies?

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 6

State Gauss's theorem in electrostatics. Using this theorem, prove that no electric field exists inside a hollow charged conducting sphere.

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 7

Explain why Gauss's law cannot be used to calculate the electric field of an electric dipole.

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 7

Explain why Gauss's law cannot be used to calculate the electric field of a charged disc.

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 7

Explain why Gauss's law cannot be used to calculate the electric field of three-point charges at the corners of a triangle.

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 8

If the radius of the gaussian surface enclosing a charge is halved, how does the electric flux through the gaussian surface change?

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 9

An electric charge

Q

is distributed on a thin conducting spherical shell uniformly. What is the electric field at the centre ? If the shape of the shell is changed will the result be different? What will happen if the spherical shell is nonconducting?

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A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 3 - Gauss' Law and its Application>EXERCISE 3>Q 10

Draw graphs showing the variations of the electric field and potential due to a spherical shell .