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GR0177 #95
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Electromagnetism$\Rightarrow$}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 $E^{'}=E_0/K$.

Alternate Solutions
 RusFortunat2015-10-16 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 $E = E_0/K$.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 ( $K$ 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 $E$).Reply to this comment herrphysik2006-10-02 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.Reply to this comment
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 $E = E_0/K$.
 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 ( $K$ 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 $E$).
 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 $\.E_{inside}$ < $\.E_0$.
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).
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.

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