32 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.we could use solid angles ig.. but this shape is pretty complex.
wouldnt that give for a disc tho
we can use it for any shape
q/4pie0 times the solid angle
would be the flux
but there should be an easier way
Yea solid angle se dekhlo ekbaar
im guessing... is the ans 5q/24e0
yeahh
solid angle ??
yup.. i calculated solid angles by inputting the coordinates
using chatgpt
cuz it's too much work
ðŸ˜
lol
the solid angle came out to be 5pi/6
approx
there should be an easier method
hnn
where's this problem from
some friends suggested making a pyramid
coaching test
oh damn
mains paper 🙂
nice coaching
frfr
Yea then the flux through 4 face will be equal cyz symmetry and flux through flat face will be q/6epsilon which is equal to q/epsilon so rearrange flux through 4 pyramidal face will be q/epsilon - q/6epsilon which is 5/6epsilon so flux through 1 pyramidal face will be 5/24epsilon
oh yeah, thanks for this
so will this work with any triangular surface? like it need not be equilateral like this one right
how symmetry
how q/6
the 4 triangular surfaces are the same
so it's symmetrical
and flat one is q/6 cuz of the cube thing
like we make 6 faces
As long as all the other triangles r symmetric or similar then yea ig (without considering for flat surface at base)
Make a cube out of 6 flat bases then flux thru each will be q/6ep
but the charge isnt in the center
we are making a triangular pyramid right
ohhh
square one
mb
m
hmm makes sens
right okay... thank you
Ohh like we did in case of cubes
Yh
nice pfp...bro started jee prep in nursery 🗿 🗿 🗿
wont we add q/24epsilon tho for that 1/4th side of the square surface
isnt 5q/24epsilon just for the sides above it
q/epsilon = q/6epsilon + 4phi where phi is required quanityt
isnt req qty phi + q/24epsilob?
make a pyramid with sq base in x-z plane
the point where the charge is somewhere on y axis
due to symmetry neighbouring 4 triangular sides of pyramid will get same flux
and square bas ka alag
add up each
+solved @integralofe^2v
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