Advertisements
Advertisements
प्रश्न
If `oint_s` E.dS = 0 over a surface, then ______.
- the electric field inside the surface and on it is zero.
- the electric field inside the surface is necessarily uniform.
- the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
- all charges must necessarily be outside the surface.
विकल्प
a and b
b and c
c and d
a and d
Advertisements
उत्तर
c and d
Explanation:
We know electric flux is proportional to the number of electric field lines and the term `oint_s` E.dS represents electric flux over the closed surface.
It means `oint_s` E.dS represents the algebraic sum of the number of flux lines entering the surface and number of flux lines leaving the surface.
If `oint_s` E.dS = 0, this means the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
From Gauss’ law, we know `oint_sE.dS = q/ϵ_0`, here q is the charge enclosed by , the closed surface. If `oint_sE.dS` = 0 then q = 0, i.e., net charge enclosed by the surface must be zero.
Hence all other charges must necessarily be outside the surface. This is because of the fact that charges outside the surface do not contribute to the electric flux.
APPEARS IN
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.
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`
Gaussian surface cannot pass through discrete charge because ____________.
The Gaussian surface ______.
The surface considered for Gauss’s law is called ______.
The Electric flux through the surface
 (i) |
 (ii) |
 (iii) |
 (iv) |
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 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.
Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then
- total flux through the surface of the sphere is `(-Q)/ε_0`.
- field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
- flux through the surface of sphere due to 5Q is zero.
- field on the surface of sphere due to –2Q is same everywhere.
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:
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?
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 ______.
