An electric dipole of dipole moment 20 × 10 − 6 C m 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

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Text Based Exercises>Q 71

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12th CBSE

IMPORTANT

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What is the use of Gaussian surface?

Important Questions on Electric Charges and Field

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IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Text Based Exercises>Q 72

How much is the electric flux through a closed surface due to a charge lying outside the closed surface?

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New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Text Based Exercises>Q 73

Two plane sheets of charge densities $+ σ$ and $- σ$ are kept in air as shown in Fig. 1.161. What are the electric field intensities at points $A$ and $B$ ?

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New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Text Based Exercises>Q 74

Two small balls, having equal positive charge

q

coulomb are suspended by two insulating strings of equal length

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metre from a hook fixed to a stand. The whole set up is taken in a satellite into space where there is no gravity. What is the angle between the two strings and the tension in each string?

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New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Text Based Exercises>Q 75

An electric dipole of dipole moment