Boundary conditions. Dielectric causes dipoles in it to direct to decrease field from inner charge q. Doing so, the q on inner shell redistributes. End result is equal field in capacitor but different charge distribution on inner shell.
Thanks a lot 😊, but I have a doubt the total capacitance can only be found if we know about the connection ( series/ parallel) so we cannot simply write C1+C2, right?
For those who are wondering about the voltage and why the signs change at the end: We know that the capacity can not be negative. The problem here is that we calculated the V(b)-V(a) which is negative. But at the end, we put V(a) -V(b) or the absolute value of the prevoius.
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I detested equations in school but when I watch the calculations in this setting, it's much easier to understand. Thank you sir.
Thank you so much. You gave me a huge help. This video was so useful and instructive!! AND MADE ME PEACEFUL
Hi !...
Where Are You From ?
Respect from the other side of Earth...!
can someone tell me why are the electric fields on both sides of the sphere equal at constant radius?
Boundary conditions. Dielectric causes dipoles in it to direct to decrease field from inner charge q. Doing so, the q on inner shell redistributes. End result is equal field in capacitor but different charge distribution on inner shell.
@@THELORDVODKA Please Explain Me ... This Thing Again ...
Thanks a lot 😊, but I have a doubt the total capacitance can only be found if we know about the connection ( series/ parallel) so we cannot simply write C1+C2, right?
what will be the capacitance of an octant of a sphere if capacitance of sphere is C?
For those who are wondering about the voltage and why the signs change at the end:
We know that the capacity can not be negative. The problem here is that we calculated the V(b)-V(a) which is negative. But at the end, we put V(a) -V(b) or the absolute value of the prevoius.
Yo got a sign error in V =( Q/2pi(e1+e2) ) * (1/a - 1/b). You said it is 1/b - 1/a....
Yes, but he needed to do integral from b to a, since the inner sphere has +Q, so consequently the result was correct..🙄
Why you find this 5:50, but I don't see you use it?
Thanks for this explanation, it's really good.
Thank you! Really helpful,
upload videos more on dielectrics and force and energy in it
Sir, there is something I have not yet understood. Why is there only radial component of electric field?
thanks for the video, but i want to ask what if the half of the sphere is vacuum(/empty)? we must take E2=0 or we need to something different?
E= e0 permitivity of the vacuum
thank you man
Thanks for all but can u plzzz complete quantum mechanics problem of Griffiths
why is the electric field the same on the two regions?
check the boundary condition at the interface. The radial field (Er) ,which is tangent to the interface, are both equal in the different two regions.
what is capital D??
Electric displacement
Thanks a lot
Is it from Griffith???
It is From in These Books :
Problems in General Physics By IE Irodov
Pathfinder Physics By Arvind Tiwari
Intro To Electrodynamics By Griffiths
Didn't Notice ..... I am Replying Two Years after you wrote this ...
thank u!