Advertisements
Advertisements
प्रश्न
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
Advertisements
उत्तर
According to Gauss' law, the electric flux through an enclosed surface is given by `oint_s vecE * dvecS = q_(enclosed)/ε_0`.
The net charge on a dipole is given by –q + q = 0, hence `q_(enclosed)` = 0
Hence the electric flux through a surface enclosing a dipole = `(-q + q)/ε_0 = q/ε_0` = 0
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?
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.
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.
The Gaussian surface ______.
Gauss' law helps in ______
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 `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.
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.
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.
A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.




