10 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.Found this (did not cross check)
https://infinitylearn.com/question-answer/two-dielectric-slabs-of-dielectric-constant-k-and--64b8e394c66212fe20be2846
Infinity Learn
Two dielectric slabs of dielectric constant K and 2K are placed ins...
The correct answer is When dielectric slabs that cover only part of the area of the capacitor plates are introduced, the charge density on the capacitor plates is no longer uniform. If the charge density on the lower half of the plates (i.e. in front of the points O and P) is σ1 and −σ1, then E0=σ12ε0 and EP=σ12(Kε0)If the charge density on the ...
Haven't done the chapter but maybe there's a mathematical mistake in second last step,
You wrote Co twice in denominator
Ah but that doesn't effect the answer tho ig.
In the solution, they have taken the charge density to be same for both the upper part and lower part of the left plate for some reason.
i think they made a typo
dv em and en should have sigma 2 further proven right since when they equated it there is sigma 2 in rhs
their end result is correct
Can you explain how?
I didn't understand the solution they provided.
Okay, I think I understand where I went wrong
+solved @iTeachChem
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