find the flux passing through hexagon.
Answers
Answer:
flux =q/ ε0
here charge is on edge so Q=q/2
so flux= q/ 2 ε0
r u an aakashian ?
Answer:
For ease of visualisation, consider the cuboid to have its longer sides vertical (with the shorter sides of length a along the x and y axes, and the longer side in the z direction). Assume also that the (point) charge is at the intersection O of the 4 body diagonals.
The length of a body diagonal b will be a√6. With O as centre, draw a spherical surface with radius r = b/2 = a√(3/2) around the point charge. All 8 corners of the cuboid will be just touching the spherical surface (our Gaussian surface of area 4πr^2).
By definition the electric flux due to any charge configuration through an area S will be the number of lines of force passing normally through S. Also, since the charge is at the centre of the spherical surface, the lines of force due to the charge will be normal to the spherical surface.
The flux through each of the 4 vertical faces will be the same and twice the flux through each horizontal face. It should be easy enough to calculate the flux through the faces.
Explanation:
it took me plenty of time to write this...please mark me the brainliest..i have explained it in detail..
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