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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.) 

संख्यात्मक
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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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Gauss’s Law
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पाठ 1: Electric Charge and Fields - EXERCISES [पृष्ठ ४३]

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

संबंधित प्रश्‍न

State and explain Gauss’s law.


Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.


A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.


State Gauss's law in electrostatics. Show, with the help of a suitable example along with the figure, that the outward flux due to a point charge 'q'. in vacuum within a closed surface, is independent of its size or shape and is given by `q/ε_0`


State Gauss’s law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the wire.


Gaussian surface cannot pass through discrete charge because ____________.


Gauss’s law is true only if force due to a charge varies as ______.

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


Gauss' law helps in ______


Five charges q1, q2, q3, q4, and q5 are fixed at their positions as shown in figure. S is a Gaussian surface. The Gauss’s law is given by `oint_s E.ds = q/ε_0`

Which of the following statements is correct?


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.

Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region

  1. the electric field is necessarily zero.
  2. the electric field is due to the dipole moment of the charge distribution only.
  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.

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.

In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.

  1. Find the critical value of y such that expansion may start.
  2. Show that the velocity of expansion is proportional to the distance from the centre.

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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