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An arbitrary surface encloses a dipole. What is the electric flux through this surface? - Physics

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प्रश्न

An arbitrary surface encloses a dipole. What is the electric flux through this surface?

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उत्तर

According to Gauss' law, the electric flux through an enclosed surface is given by `oint_s vecE * dvecS = q_(enclosed)/ε_0`.

The net charge on a dipole is given by –q + q = 0, hence `q_(enclosed)` = 0

Hence the electric flux through a surface enclosing a dipole = `(-q + q)/ε_0 = q/ε_0` = 0

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Gauss’s Law
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अध्याय 1: Electric Charges And Fields - MCQ I [पृष्ठ ५]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 12
अध्याय 1 Electric Charges And Fields
MCQ I | Q 1.14 | पृष्ठ ५

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The Electric flux through the surface


(i)

(ii)

(iii)

(iv)

If `oint_s` E.dS = 0 over a surface, then ______.

  1. the electric field inside the surface and on it is zero.
  2. the electric field inside the surface is necessarily uniform.
  3. the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
  4. all charges must necessarily be outside the surface.

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  3. the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
  4. the work done to move a charged particle along a closed path, away from the region, will be zero.

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  2. field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
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  4. field on the surface of sphere due to –2Q is same everywhere.

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