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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the Figure. What is the magnitude of the electric flux through the square? (Hint: Think of the - Physics

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प्रश्न

A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the Figure. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.) 

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उत्तर

Based on the illustration, we can assume that the given square represents one side of a cube with an edge length of 10 cm.

The cube encloses the +10 µC charge because it is located 5 cm above the square's center. According to Gauss's theorem, cube-linked electric flow is

Φ = `q/ε_0`

= `(10 xx 10^-6)/(8.85 xx 10^-12)`

= 1.13 × 106 N m2 C−1

Consequently, the electromotive force acting on the square

Φsq = `"Φ"/6`

= `1.13/6 xx 10^6`

or Φsq = 1.9 × 105 N m2 C−1

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पाठ 1: Electric Charge and Fields - EXERCISES [पृष्ठ ४३]

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एनसीईआरटी Physics Part 1 and 2 [English] Class 12
पाठ 1 Electric Charge and Fields
EXERCISES | Q 1.17 | पृष्ठ ४३

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Gauss's law is valid for ______.

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The surface considered for Gauss’s law is called ______.


Gauss' law helps in ______


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.

If there were only one type of charge in the universe, then ______.

  1. `oint_s` E.dS ≠ 0 on any surface.
  2. `oint_s` E.dS = 0 if the charge is outside the surface.
  3. `oint_s` E.dS could not be defined.
  4. `oint_s` E.dS = `q/ε_0` if charges of magnitude q were inside the surface.

Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then

  1. total flux through the surface of the sphere is `(-Q)/ε_0`.
  2. field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
  3. flux through the surface of sphere due to 5Q is zero.
  4. field on the surface of sphere due to –2Q is same everywhere.

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In finding the electric field using Gauss law the formula `|vec"E"| = "q"_"enc"/(epsilon_0|"A"|)` is applicable. In the formula ε0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?


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