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
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2.
- Find the charge on the sphere.
- What is the total electric flux leaving the surface of the sphere?
Define Electric Flux.
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?
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?
A charged particle q is placed at the centre O of cube of length L (A B C D E F G H). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is ______.
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 ______.
What is the SI unit of electric flux?
For an open surface, which statement correctly describes the area vector convention?
For a closed surface, which convention must be followed when assigning the area vector?
What does negative electric flux indicate?
