1*edited*
Can't we solve question 1 without solid angle stuff...?
1.A point charge is placed at the origin.Calculate electric flux through the open hemispherical surface:......
{x-a}2+y2+z2=a2;x≥a
Ans.q/2ε0*{1-1/√2)
2.A point charge is located on the axis of a disc of radius R at a distance b from the plane of the disc .Show that if one fourth of flux pass through the disc then ,R=b.
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10 Answers
49we can use the concpt that the net flux= 0
so the flux through the base part of hemisphere=flux through the spherical part
is this not? verify...
11Ans 2) I am getting R = √3 b LOL [89]
Angle subtended by disc at Q =W = 2 Π(1 - cos α)
W = 2 Î [1 - b√b2 +R2 ]
Flux thru the disc = Q W4Πε
Since one fourth of flux pass through the disc,
therefore, Q W4Πε = Q 4 ε
Thus, R = √3 b
1Well its a question is from Arihant,,... (pg 170 ,vol4)
then confirm in the words "question is wrong"