GREPhysics.NET: GR0177 #95

GR0177 #95

Problem

GREPhysics.NET Official Solution

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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

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 $E = E_0/K$ . Reply to this comment
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. Reply to this comment
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 $E$ ). Reply to this comment
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. Reply to this comment

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

Reply to this comment

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

Reply to this comment

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?

Reply to this comment

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 $\.E_{inside}$ < $\.E_0$ .

Reply to this comment

Mexicana
2007-10-02 17:52:53

Reply to this comment

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.

Reply to this comment

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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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