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 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?
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 for magnetism. Explain its significance.
Answer the following question.
State Gauss's law for magnetism. Explain its significance.
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`
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 ____________.
The Gaussian surface ______.
Gauss' law helps in ______
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.
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 the diagram, the total electric flux through the closed surface ‘S’ is:
[Given q = charge ε0 = permittivity of free space]
