Advertisements
Advertisements
प्रश्न
If there were only one type of charge in the universe, then ______.
- `oint_s` E.dS ≠ 0 on any surface.
- `oint_s` E.dS = 0 if the charge is outside the surface.
- `oint_s` E.dS could not be defined.
- `oint_s` E.dS = `q/ε_0` if charges of magnitude q were inside the surface.
पर्याय
a and d
a and c
b and d
c and d
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.
APPEARS IN
संबंधित प्रश्न
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 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 for magnetism. Explain its significance.
Answer the following question.
State Gauss's law for magnetism. Explain its significance.
Gauss' law helps in ______
The Electric flux through the surface
 (i) |
 (ii) |
 (iii) |
 (iv) |
Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region
- the electric field is necessarily zero.
- the electric field is due to the dipole moment of the charge distribution only.
- the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
- the work done to move a charged particle along a closed path, away from the region, will be zero.
A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.
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:
The algebraic form of Gauss's Law is represented as:
The integral form of Gauss's Law is given by:
In the context of Gauss's Law, what does the area element vector dS represent?
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)?
Gauss's Law is equivalent to:
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:
