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If we have two parallel plate capacitors in parallel, do the electric fields between their plates have to be the same? What happens if we have a coaxial cable of length $L$, which looks like this:

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One part of it (length $X$) is filled with a dielectric of relative permittivity $\epsilon_r$. Are the two electric fields $E_1$ and $E_2$ between the inner and outer electrodes the same? If we look at the cable as two separate cylindrical capacitors of lengths $X$ and $L-X$, are those capacitors then connected in series or parallel in this case?

Edit: Electric field using Gauss's law (in case $X=\frac{L}{2}$).

Since the two cylindrical capacitors are in parallel, $E_1$ and $E_2$ are the same, so we know, by Gauss's law, that: $$\oint_S{DdS=Q}$$ $$\epsilon_0E2{\pi}r\frac{l}{2}+\epsilon_0\epsilon_rE2{\pi}r\frac{l}{2}=Q$$ $$\epsilon_0E(1+\epsilon_r){\pi}rl=Q$$ $$E=\frac{Q}{{\pi}rl\epsilon_0(1+\epsilon_r)}$$ I'm not entirely sure if I correctly solved the integral.

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it may be considered as connected in parallel and electric field is same. since energy stored = 1/2 e°E^2 in upper part and 1/2 eE^2 in lower charge stored by 2 are different lower stores more charge so stores more energy

both have same potential difference Q= CV Q' = C' V capacity of both are different. using gauss law electric field in both is E.S=Q/e°. in second case E'.S= Q'/e°er where extra charge is cancelled by extra 'epsilon r'. so E is same in both cases.

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  • $\begingroup$ Then the total charge Q stored in those capacitors is not equally distributed between them. Does that result in different electric fields? If the fields are different, how can the voltage be the same? $\endgroup$ – A6SE Oct 24 '15 at 14:18
  • $\begingroup$ One more thing bothers me, if electric fields are the same, how do we find the electric field? I posted how I did it using Gauss's law, but I'm not entirely sure what Gaussian surface I should take, because the charge is not distributed equally across the whole cable. $\endgroup$ – A6SE Oct 24 '15 at 18:41

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