Advertisements
Advertisements
प्रश्न
A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.
पर्याय
`Q/epsi_0`
`Q/(6epsi_0)`
`Q/(8epsi_0)`
`Q/(3epsi_0)`
Advertisements
उत्तर
A charge Q is placed at the centre of a cube. The electric flux through one of its faces is `underlinebb(Q/(6epsi_0))`.
Explanation:
Assuming the cube as a Gaussian surface, the total flux passing through the cube of side l can be given as,
`phi = Q/epsi_0`
Flux through each face of the cube: As the charge is placed symmetrically to each face of the cube, thus electric flux passing through each face is equal.
∴ Electric flux passing through each face,
`phi^' = phi/6 = Q/(6epsi_0)`
APPEARS IN
संबंधित प्रश्न
Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.
Gauss' law helps in ______
The Electric flux through the surface
![]() (i) |
![]() (ii) |
![]() (iii) |
![]() (iv) |
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.
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?
The integral form of Gauss's Law is given by:
In the context of Gauss's Law, what does the area element vector dS represent?
The electric field due to a uniformly charged infinite plane sheet with surface charge density σ is:
In a non-uniform electric field given by \[\vec{E} = \alpha x^{1/2} \hat{i}\] with \[\alpha = 800 \text{ N C}^{-1}\text{m}^{-1/2}\], the flux through a cube of side a = 0.1 m oriented with faces perpendicular to the x-axis is approximately:




