Advertisements
Advertisements
प्रश्न
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.
विकल्प
b and d
a and c
b and d
c and d
Advertisements
उत्तर
c and d
Explanation:
From Gauss’ law, we know `oint_s` E.dS = `q_(enclosed)/ε_0` in left side equation.
The electric field is due to all the charges present both inside as well as outside the Gaussian surface. Hence if `q_(enclosed)` = 0, it cannot be said that the electric field is necessarily zero.
If there are various types of charges in a region and total charge is zero, the region may be supposed to contain a number of electric dipoles.
Therefore, at points outside the region (maybe anywhere w.r.t. electric dipoles), the dominant electric field `∞ 1/r^3` for large r.
The electric field is conservative, work done to move a charged particle along a closed path, away from the region will be zero.
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?
State Gauss’s law for magnetism. Explain its significance.
Answer the following question.
State Gauss's law for magnetism. Explain its significance.
The Gaussian surface ______.
The surface considered for Gauss’s law is called ______.
Gauss' law helps in ______
The Electric flux through the surface
![]() (i) |
![]() (ii) |
![]() (iii) |
![]() (iv) |
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.
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.
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
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.
In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.
- Find the critical value of y such that expansion may start.
- Show that the velocity of expansion is proportional to the distance from the centre.
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?
A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.




