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

The Electric flux through the surface (i) (ii) (iii) (iv)

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प्रश्न

The Electric flux through the surface


(i)

(ii)

(iii)

(iv)

पर्याय

  • in Figure (iv) is the largest.

  • in Figure (iii) is the least.

  • in Figure (ii) is same as Figure (iii) but is smaller than Figure (iv)

  • is the same for all the figures.

MCQ
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उत्तर

is the same for all the figures.

Explanation:

According to Gauss’ law of electrostatics, the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity,

i.e., `phi = (Q_(enclosed))/ε_0`

Thus, electric flux through a surface doesn’t depend on the shape, size or area of a surface but it depends on the amount of charge enclosed by the surface.

In the given figures the charge enclosed is the same that means the electric flux through all the surfaces should be the same.

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पाठ 1: Electric Charges And Fields - MCQ I [पृष्ठ २]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
पाठ 1 Electric Charges And Fields
MCQ I | Q 1.03 | पृष्ठ २

संबंधित 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.)


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?


A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.


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


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 ______.

Gauss's law is valid for ______.

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


Gauss' law helps in ______


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.

If the total charge enclosed by a surface is zero, does it imply that the elecric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.


The region between two concentric spheres of radii a < b contain volume charge density ρ(r) = `"c"/"r"`, where c is constant and r is radial- distanct from centre no figure needed. A point charge q is placed at the origin, r = 0. Value of c is in such a way for which the electric field in the region between the spheres is constant (i.e. independent of r). Find the value of c:


A charge of +5 μC is placed at the centre of two concentric spheres of radii r1 = 3 cm and r2 = 5 cm. The ratio of the flux through sphere of radius r1 to that through sphere of radius r2 will be ______.


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