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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.

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प्रश्न

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.

थोडक्यात उत्तर
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उत्तर

ccording to Gauss's Law, flux passing through any closed surface is equal to `1/∈_0` tin the charge enclosed by that surface.

`=> phi = "q"/∈_0,`

where ϕ is the flux through the closed surface and q is the charge enclosed by that surface.
The charge is placed at the centre of the cube and the electric field is passing through the six surfaces of the cube.vSo, we can say that the total electric flux passes equally through these six surfaces .
Thus, flux through each surface,

`phi ′ = "Q"/(6∈_0)`

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पाठ 30: Gauss’s Law - Exercises [पृष्ठ १४१]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 30 Gauss’s Law
Exercises | Q 5 | पृष्ठ १४१

संबंधित प्रश्‍न

State and explain Gauss’s law.


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.)


A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube?


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


State Gauss’s law for magnetism. Explain its significance. 


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 ____________.


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


The Electric flux through the surface


(i)

(ii)

(iii)

(iv)

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.

Which of the following is NOT a property of a Gaussian surface?


For a Gaussian surface to be convenient for calculations, the electric field should be:


According to Gauss's Law, the total electric flux through a closed surface depends on:


The algebraic form of Gauss's Law is represented as:


The integral form of Gauss's Law is given by:


For an infinite line charge with linear charge density λ, the electric field at distance r from the wire is:


The electric field due to a uniformly charged infinite plane sheet with surface charge density σ is:


Gauss's Law is equivalent to:


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