Electrostats doubt- Field just below a hemispherical surface
Isn't the disc electric field formula applicable for this q? Derived by using solid angle. Could someone show their working? I have a couple of doubts.
45 Replies
ik this doesnt look like a hemisphere but i hope u get what i mean
@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.I'm guessing, but is it D?
sorry no
Mb A
i misread it mb
nope i thought that too
should i tell?
ok, then let me actually try
Idk why I even said D
A is the more likely one
yeah cuz if it's close it acts like an infinite sheet
but idk why that is not it
but A isnt correct right?
nope
alrr
at least acc to the key
key says C?
yeah
damn
if u use the disc wala formula you get the same exp except that it has a - sign and the ans has a + sign
and im assuming the hemisphere doesnt have a base ie it's open
i was trying with gauss law but couldnt get much
ping me before closing if u get ans
okay sure
i think its D too
by the sheet logic
there will be 2 sheets
a better img i found online (same q)
and i think the flat part is open
the q doesnt explicitly mention anything like that
integrate this from 0 to pi/2
gives D if my calc is correct
isnt there smtg simpler than this? like maybe solid angles
No it doesnt
I will send the calculation tomorrow i dont have a pen and its hard writing on laptop
Thats for discs right
these are rings
If this ain't correct imma kms
There might be just 1 error that maybe i took d exactly as zero and didnt use it as a limit so it might cause some issues
Ah I've seen this. This is painfully done using polar ðŸ˜
But yeah the answer is b for this case.
c occurs for the case when the particle is at a distance R opposite to the center iirc.
This is the trickiest part in the question
Because d is tending to zero, the field by the first ring (when theta is zero) is not zero, but sigma/2epsilon
the key says it's c tho
wont it be (root2-1) instead of root2+1
We have missed the extra sigma/2epsilon component of E-field that's why we add it to the obtained one.
Yeah my bad. c for this one and b for the other case.
This gives c.
okay.. thanks a bunch.. i'll need some time to check and understand it. will close the post after that
@hardcoreisdead u asked me to ping right...
Cool u can find that field by setting limit as d tenda to zero when using the formula for field by disc right? Tha gives sigma/2epsilon
Yep.
Alr mate ty
ok so to sum it up field at A is option C and at point B is option B
why is method of integration shared by hithav wrong
Because of the extra sigma/2epsilon component of electric field that we have taken to be zero in integration has to be added to obtain net field.
+solved @SirLancelotDuLac
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