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.
पर्याय
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
State and explain Gauss’s law.
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 ≤ ∞.
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`
Gaussian surface cannot pass through discrete charge because ____________.
The surface considered for Gauss’s law is called ______.
Which of the following statements is not true about Gauss’s law?
Gauss' law helps in ______
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?
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?
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 ______.
