Advertisements
Advertisements
प्रश्न
Find out the outward flux to a point charge +q placed at the centre of a cube of side ‘a’. Why is it found to be independent of the size and shape of the surface enclosing it? Explain.
Advertisements
उत्तर
Let a cube of side a enclose charge +q at its centre.
Because the electric flux through the square surface is `phi=q/(6in_0)`the square surfaces of cube are six. Hence, according to Gauss’s theorem in electrostatics, the total outward flux due to a charge +q of a cube is
`phi=6xx(q/(6in_0))=q/in_0`
The result shows that the electric flux passing through a closed surface is proportional to the charge enclosed. In addition, the result reinforces that the flux is independent of the shape and size of the closed surface.
APPEARS IN
संबंधित प्रश्न
Define electric flux.
Define Electric Flux.
A small plane area is rotated in an electric field. In which orientation of the area, is the flux of the electric field through the area maximum? In which orientation is it zero?
Following Figure (a) shows an imaginary cube of edge L/2. A uniformly charged rod of length (L) moves towards the left at a small but constant speed `nu.` At t = 0, the left end just touches the centre of the face of the cube opposite it. Which of the graphs shown in the figure (b) represents the flux of the electric field through the cube as the rod goes through it?

If the flux of the electric field through a closed surface is zero,
(a) the electric field must be zero everywhere on the surface
(b) the electric field may be zero everywhere on the surface
(c) the charge inside the surface must be zero
(d) the charge in the vicinity of the surface must be zero
The following figure shows a closed surface that intersects a conducting sphere. If a positive charge is placed at point P, the flux of the electric field through the closed surface

A uniform electric field of intensity 400 N/C, exists in a certain region. How much flux will cross a given area of 10 cm2 in this region, if the area vector is inclined at 60° to the direction of the field?
The electric field in a region is given by `bar"E" = 4hat"i" + 10hat"j"` N/C. The flux of this field through a square of 10 cm on a side whose plane is parallel to the XZ plane.
A hollow sphere of radius R has a point charge q at its centre. Electric flux emanating from the sphere is X. How will the electric flux change, if at all, when charge q is replaced by an electric dipole?
