See equation 2.22. Integrating from a to b is the same as minus the integral from b to a; and vice versa. So integrate from the bottom l=0 to l=a in slab 2, then from l = a to l = 2a in slab 1. The angle between E and dl is 180 in both slabs.
@@subhayumukherjee2893No.Its a difference but if you look at the theory Vb-Va=Integral of -E•dl.He explained how he did it to skip the whole process of integration.
Hello I have a question regarding using Gauss for Displacement. I understood your explanation but in the solutions for the book, they just have DxA = sigmaxA therefore D=sigma. How does that make sense?
Why at the first when you find the displacement you consider that D are equal at both sides of the gaussian surface... I mean on one side there is a vacume and on the other on the other its dielectric!!! Shouldn't we put D1 and D2 ??
The electric dispacement is only influencent by "outer Charges", it basically ignores the dielectric, which you can see when looking at the divergence of D, which is jus the outer charge density.
so clear and precise...thanks a lot...u made my day!! keep posting such useful videos!
Very neat and complete explanation, it was really useful. Thank you very much
you deserve way more subscribers!! Thank You SO MUCH!!!! :D
Really pedagogical. Congratulations Jordan Bell
This is a must watch!!
Thank you so much! You have helped so much! Please keep posting videos!
Thank you so much, keep posting !
I don't understand what is going on with the negative sign during the integration to get V.. why are both Ea terms positive?
Same
See equation 2.22. Integrating from a to b is the same as minus the integral from b to a; and vice versa. So integrate from the bottom l=0 to l=a in slab 2, then from l = a to l = 2a in slab 1. The angle between E and dl is 180 in both slabs.
@@brianholloway7468but there should be a minus sign innit? In the question it was asked to find the potential 'difference'
@@subhayumukherjee2893No.Its a difference but if you look at the theory Vb-Va=Integral of -E•dl.He explained how he did it to skip the whole process of integration.
Hello I have a question regarding using Gauss for Displacement. I understood your explanation but in the solutions for the book, they just have DxA = sigmaxA therefore D=sigma. How does that make sense?
Great video!
U are awesome ...
you are awesome!
4.19 please
so useful
4.26?
Why at the first when you find the displacement you consider that D are equal at both sides of the gaussian surface... I mean on one side there is a vacume and on the other on the other its dielectric!!! Shouldn't we put D1 and D2 ??
Because the free charge is the same at both but diffrent in signe
The electric dispacement is only influencent by "outer Charges", it basically ignores the dielectric, which you can see when looking at the divergence of D, which is jus the outer charge density.
@@maxmoeller3597 But Displacement is Electric field + Polarisation so inside a polarised dielectric, shouldn't it be different?
Thanks
Thanks