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As shown in the figure, a charge q is placed at the centre of the open end of a cylinder that is open at one end and closed at the other. The total flux emerging from the surface of the cylinder is

Asked in RS Academy GUJCET booklet · Gauss's law and enclosed charge

Figure: Gauss's law and enclosed charge
Answer: (3) q/(2ε₀)

Step-by-step solution

Idea: close the cylinder with an imaginary flat lid across the open end. The charge then sits on the lid, in its plane.

A charge lying on a flat part of a closed surface has exactly half of its field lines going into the enclosed side, so the flux through the closed surface is 1/2×q/(ε₀).

Through the imaginary lid itself the flux is zero: at every point of the lid the field of q lies along the lid, parallel to its surface.

So all of that half passes through the real cylinder - its curved wall and its closed base.

Total flux through the cylinder =q/(2ε₀), whatever the length and radius of the cylinder.

Why the other options are wrong

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