Advertisements
Advertisements
प्रश्न
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
Advertisements
उत्तर
According to Gauss's law, flux through a closed surface is given by
`phi = q/ε_0`
Here, q is the charge enclosed by the Gaussian surface.
Since, on increasing the radius of the Gaussian surface, charge q remains unchanged, the flux through the spherical Gaussian surface will not be affected when its radius is increased.
संबंधित प्रश्न
Given the electric field in the region `vecE=2xhati`, find the net electric flux through the cube and the charge enclosed by it.
Two charges of magnitudes −3Q and + 2Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘5a’ with its centre at the origin?
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’?
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?
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?
A charge 'Q' µC is placed at the centre of a cube. The flux through one face and two opposite faces of the cube is respectively ______.
The electric flux through the surface ______.
 |
 |
 |
 |
| (i) | (ii) | (iii) | (iv) |
The SI unit of electric flux is ______.
In a region of space having a uniform electric field E, a hemispherical bowl of radius r is placed. The electric flux Φ through the bowl is:
