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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.
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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.
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संबंधित प्रश्न
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Given a uniform electric filed \[\vec{E} = 4 \times {10}^3 \ \hat{i} N/C\]. Find the flux of this field through a square of 5 cm on a side whose plane is parallel to the Y-Z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?
The total flux through the faces of the cube with side of length a if a charge q is placed at corner A of the cube is ______.
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- A: a corner of the cube.
- B: mid-point of an edge of the cube.
- C: centre of a face of the cube.
- D: mid-point of B and C.
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