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Electromagnetism}Dielectric

No real calculations involved here. Since E\propto 1/\epsilon, if one adds in a dielectric then \epsilon = K\epsilon_0. Since E_0\propto 1/\epsilon_0, this means that E^{'}=E_0/K.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
RusFortunat
2015-10-16 22:17:54
What if we simply use the fact that D_{1n} = D_{2n} on the border, or \\eps_1 E_0 = \\eps_2 E. Applying this to our conditions we get E = E_0/K.Alternate Solution - Unverified
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.Alternate Solution - Unverified
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 ( K 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 E).Alternate Solution - Unverified
herrphysik
2006-10-02 18:42:02
Just use the equation (4.35 in Griffiths) E=E_{vac}/\epsilon_r where \epsilon_r=\epsilon/\epsilon_0 is the relative permittivity. Plug in the given \epsilon=K\epsilon_0 to get the correct answer.Alternate Solution - Unverified
Comments
RusFortunat
2015-10-16 22:17:54
What if we simply use the fact that D_{1n} = D_{2n} on the border, or \\eps_1 E_0 = \\eps_2 E. Applying this to our conditions we get E = E_0/K.
RusFortunat
2015-10-16 22:20:53
D: in preview it was more nice
Alternate Solution - Unverified
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
NEC
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.
f4hy
2009-11-07 00:27:13
Doesn't (B) satisfy those limits as well?
Alternate Solution - Unverified
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 ( K 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 E).
Moush
2010-09-18 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 \.E_{inside} < \.E_0.
Alternate Solution - Unverified
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 (K is dimensionless).NEC
herrphysik
2006-10-02 18:42:02
Just use the equation (4.35 in Griffiths) E=E_{vac}/\epsilon_r where \epsilon_r=\epsilon/\epsilon_0 is the relative permittivity. Plug in the given \epsilon=K\epsilon_0 to get the correct answer.Alternate Solution - Unverified

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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

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