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GR0177 #95
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Electromagnetism}Dielectric

No real calculations involved here. Since , if one adds in a dielectric then . Since , this means that .  Alternate Solutions
 RusFortunat2015-10-16 22:17:54 What if we simply use the fact that on the border, or . Applying this to our conditions we get .Reply to this comment fcarter2008-10-13 17:16:57 Another one where limits are the fast way. For K->1, E->E. For K->inf, E->0. Only answer that satisfies is A.Reply to this comment Mexicana2007-10-02 17:57:24 Another way to increase your chances of getting it right is by elimination of choices B, C and D since using dimensional analysis only choice A and E have the correct dimensions of electric field ( is dimensionless). If with this you also remember the fact that a dielectric under an applied electric field, induces an internal field so that the total field inside decreases, then only choice A remains (as this is the only one which would decrease the original field ).Reply to this comment herrphysik2006-10-02 18:42:02 Just use the equation (4.35 in Griffiths) where is the relative permittivity. Plug in the given to get the correct answer.Reply to this comment RusFortunat
2015-10-16 22:17:54
What if we simply use the fact that on the border, or . Applying this to our conditions we get .
 RusFortunat2015-10-16 22:20:53 D: in preview it was more nice johnVay
2013-10-16 19:06:28
dielectics (like in capacitors) increasing rearranging their internal charges to oppose
the E field. in the limit they are conductors where they rearrange perfectly.

the order:
vacuum - dielectic - conductor

dielectrics always scaling the electric field by kappa. this leaves a and b, you need a little more to recognize that B has an extra factor of epsilon fcarter
2008-10-13 17:16:57
Another one where limits are the fast way. For K->1, E->E. For K->inf, E->0. Only answer that satisfies is A.
 f4hy2009-11-07 00:27:13 Doesn't (B) satisfy those limits as well? Mexicana
2007-10-02 17:57:24
Another way to increase your chances of getting it right is by elimination of choices B, C and D since using dimensional analysis only choice A and E have the correct dimensions of electric field ( is dimensionless). If with this you also remember the fact that a dielectric under an applied electric field, induces an internal field so that the total field inside decreases, then only choice A remains (as this is the only one which would decrease the original field ).
 Moush2010-09-18 15:55:06 Choice C has the same units as A and E so you can't eliminate it based only on dimensional analysis, but it's intuitive that < . Mexicana
2007-10-02 17:52:53
Another way to increase your chances of getting it right is by elimination of choices B, C and D since using dimensional analysis only choice A and E have the correct dimensions of electric field ( is dimensionless). herrphysik
2006-10-02 18:42:02
Just use the equation (4.35 in Griffiths) where is the relative permittivity. Plug in the given to get the correct answer.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$