Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
ccording to Gauss's Law, flux passing through any closed surface is equal to `1/∈_0` tin the charge enclosed by that surface.
`=> phi = "q"/∈_0,`
where ϕ is the flux through the closed surface and q is the charge enclosed by that surface.
The charge is placed at the centre of the cube and the electric field is passing through the six surfaces of the cube.vSo, we can say that the total electric flux passes equally through these six surfaces .
Thus, flux through each surface,
`phi ′ = "Q"/(6∈_0)`
APPEARS IN
संबंधित प्रश्न
State and explain Gauss’s law.
Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.
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 ____________.
q1, q2, q3 and q4 are point charges located at points as shown in the figure and S is a spherical gaussian surface of radius R. Which of the following is true according to the Gauss' law?

The surface considered for Gauss’s law is called ______.
Which of the following statements is not true about Gauss’s law?
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.
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.
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.
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:
A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.
