conceptual lag in flux
according to gauss law flux through this random body should be q1/epsilon. q2 has no contribution in the flux since the electric field lines that enter the body also exit it thus flux is zero (source :https://youtu.be/ZR9pNx90YRU?si=QR84K-VpzGPqruSZ&t=3981)
Mohit Tyagi
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@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.if we apply a similar logic to this ques
it becomes zero ??
Uniform constant electric field?
Yeah, zero
ans is 2ERH tho
in phy galaxy
This is a uniform electric field right? So the flux will be area of projection*Electric field.
Wait I had a visualization for this...
Draw a line of electric field cutting the shape at two points. Now the area of projection of small areas are proportional to the distance^2 (Because the radius of curvature is from point to that small area), while the electric field is proportional to 1/r^2
This cancels out and hence there is no flux by the outside charge.
The same doesn't occur in uniform electric field.
is the logic i gave incorrect? and why is it not applicable in the cylinder ques
We can't apply a similar logic here because of the lack of proportionality to 1/r^2.
Wait a min I read the question wrong.
waiting
Yeah nvm it should be zero, I'm dumb 😅
What is the source, if I may ask?
ig phy galaxy high then
phy galaxy
It should have been a half cylinder here.
Otherwise by gauss law there is charge inside which is not the case.
yes that makes more sense
btw
can an objects own electric field give it flux
yes ig
"give it flux"?
Charged objects enclised in a gaussian surface do give flux.
like if theres a charged disc , it has some field lines. can those field lines give it flux
Like a gaussian surface enclosing the disc? Yep.
the disc only
acc to me yes
cuz well theres enclosed charge
Gaussian surfaces are imaginary.
oh nvm i was thinking smthing else
what if we tilt the entire cylinder by angle theta wrt vertical
keeping dirn of E same
If it's a full cylinder again zero.
If its a half cylinder, area of projection * magnitude of E-field
consider a closed cylinder , and half it
given radius R and height H
Does this help?
A half cylinder with no flat face opposing the curved one, to be precise. Needs to be open.
In fact, that's our basis for applying the projection
what i wanted to know was if we tilt the cylinder how do we consider tha flux through cap of cylinder
yeahhh
+solved @Opt @SirLancelotDuLac
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