Advertisements
Advertisements
प्रश्न
What is the electric flux through a cube of side 1 cm which encloses an electric dipole?
उत्तर
As per the Gauss's law of electrostatics, electric flux through a closed surface is given by
`phi_E=ointvecE.vec(ds)=Q/epsilon_0 `
where
E = Electrostatic field
Q = Total charge enclosed by the surface
∮E=∮E→.ds→=0ε0=0" role="presentation" style="position: relative;">∮E=∮E→.ds→=0ε0=0
∈0 = Absolute electric permittivity of free space
In the given case, cube encloses an electric dipole. Therefore, the total charge enclosed by the cube is zero, i.e. Q = 0.
Therefore, from (i), we have
`phi_E=ointvecE.vec(ds)=0/epsilon_0=0`
APPEARS IN
संबंधित प्रश्न
Write its (electric flux.) S.I unit.
"The outward electric flux due to charge +Q is independent of the shape and size of the surface which encloses is." Give two reasons to justify this statement.
Given the electric field in the region `vecE=2xhati`, find the net electric flux through the cube and the charge enclosed by it.
Figure shows three point charges +2q, −q and + 3q. Two charges + 2q and −q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’?
Mark the correct options:
The electric field in a region is given by `vec"E"` = 5 `hatk`N/C. Calculate the electric flux Through a square of side 10.0 cm in the following cases
- The square is along the XY plane
- The square is along XZ plane
- The normal to the square makes an angle of 45° with the Z axis.
An electric charge q is placed at the centre of a cube of side a. The electric flux on one of its faces will be ______.
An electric charge q is placed at the center of a cube of side ℓ. The electric flux on one of its faces will be ______.
The electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 20 V-m. The flux over a concentric sphere of radius 20 cm will be ______.
