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

If there were only one type of charge in the universe, then ______. ∮s E.dS ≠ 0 on any surface. ∮s E.dS = 0 if the charge is outside the surface. ∮s E.dS could not be defined. ∮s E.dS = εqε0

Advertisements
Advertisements

प्रश्न

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.

विकल्प

  • a and d

  • a and c

  • b and d

  • c and d

MCQ
रिक्त स्थान भरें
Advertisements

उत्तर

b and d

Explanation:

From Gauss’ law, we know `oint_s` E.dS = `q/ε_0`, here q is the charge enclosed by the closed surface. If `oint_s` E.dS =  0 then q = 0, i.e., net charge enclosed by the surface must be zero. If the charge is outside the surface, then charge enclosed by the surface is q = 0 and thus, `oint` E.dS = 0.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Electric Charges And Fields - MCQ I [पृष्ठ ४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
अध्याय 1 Electric Charges And Fields
MCQ I | Q 1.1 | पृष्ठ ४

संबंधित प्रश्न

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.


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 ____________.


q1, q2, q3 and q4 are point charges located at points as shown in the figure and S is a spherical gaussian surface of radius R. Which of the following is true according to the Gauss' law?


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


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?


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.

In the diagram, the total electric flux through the closed surface ‘S’ is:

[Given q = charge ε0 = permittivity of free space]


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 ______.


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:


What is the SI unit of permittivity of free space (ε₀)?


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:


For a uniformly charged thin spherical shell, what is the electric field at any point inside the shell (r < R)?


In a non-uniform electric field given by \[\vec{E} = \alpha x^{1/2} \hat{i}\] with \[\alpha = 800 \text{ N C}^{-1}\text{m}^{-1/2}\], the flux through a cube of side a = 0.1 m oriented with faces perpendicular to the x-axis is approximately:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×