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Question

A charge Q is present inside a cube, then find out flux due to cube and flux due to each face of a cube?


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Solution

Step 1: Given data

Charge inside the cube = Q

The dielectric constant of the vacuum = ε0

Faces of a cube = 6

Step 2: Concept applied

Gauss's law

Gauss's law for electricity states that the electric flux ϕ across any closed surface is proportional to the net electric charge qenclosed by the surface.

i.e., ϕ=qε0

  1. Flux (ϕ) is defined as the electric field flowing out through a specific area.
  2. Consider a surface element dS=n^dS in an electric field E , where n^ is the outward unit vector normal to the surface element.
  3. The quantity dϕ=E.dS=EcosθdS is called the flux of E through dS.
  4. If we consider that qε0 number of lines of force emanate isotropically from a point charge q then the idea of electric flux becomes meaningful.
  5. If we use this picture of an electric field then dϕ=E.dS becomes equal to the number of lines of force passing through the area dS.
  6. The flux of E over any arbitrary surface S is given by the integral Φ=dϕ=SE.dS.

Step 3: Calculation and conclusion

  1. Since Charge Q is present inside the cube, the flux through the cube will beϕ=qε0
  2. The flux will be one-sixth of the total flux=16(since a cube has 6 faces)
  3. Flux through each face will be = q6ε0

Hence, flux due to cube is ϕ=qε0 and flux due to each face of a cube is q6ε0


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