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प्रश्न
A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube?
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उत्तर
By using Gauss’s Law.
It is given as
`Phi = oint vecE*dvecs = q/in_0`
Now, the flux passing through all the six surfaces would be
`Phi = 6phi =q/in_0`
And the flux passing through each surface would be
`phi = q/(6in_0)`
संबंधित प्रश्न
State and explain Gauss’s law.
A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.
A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.
State Gauss's law in electrostatics. Show, with the help of a suitable example along with the figure, that the outward flux due to a point charge 'q'. in vacuum within a closed surface, is independent of its size or shape and is given by `q/ε_0`
The surface considered for Gauss’s law is called ______.
Which of the following statements is not true about Gauss’s law?
If `oint_s` E.dS = 0 over a surface, then ______.
- the electric field inside the surface and on it is zero.
- the electric field inside the surface is necessarily uniform.
- the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
- all charges must necessarily be outside the surface.
Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then
- total flux through the surface of the sphere is `(-Q)/ε_0`.
- field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
- flux through the surface of sphere due to 5Q is zero.
- field on the surface of sphere due to –2Q is same everywhere.
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
