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An isosceles right-angled triangle of leg-length d lies in a horizontal plane. The right angle is at one corner. A point charge q is placed at a vertical distance d directly above that right-angle corner. Find the electric flux through the triangular surface.
- q / (6*epsilon0)
- q / (18*epsilon0)
- q / (24*epsilon0)
- q / (48*epsilon0)
Correct answer: q / (48*epsilon0)
Solution
Place q at a corner of a cube of side d. By symmetry, 8 such cubes fill all space around q, so each cube gets flux q/(8*epsilon0). The 3 faces adjacent to the charge have zero flux (field parallel to them at the edges). The other 3 non-adjacent faces share the cube's flux equally: each gets q/(24*epsilon0). The bottom face (the d x d horizontal square) receives q/(24*epsilon0). The isosceles right triangle of side d is exactly half of this square, so its flux = q/(48*epsilon0).
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