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**[SOLVED] Griffiths Problem 4.18**

**1. Homework Statement**

This question refers to Griffiths E and M book.

Obviously we are supposed to use equation 4.23 here. But for that to be of any use, we need to assume that the electric displacement is orthogonal to the plates. We are told that the medium is linear, but that only means that [itex]\vec{D} = \epsilon \vec{E} [/itex] where [itex]\vec{E}[/itex] is the

*total*electric field. What we need to assume is that [itex]\vec{D} [/itex] is parallel to [itex] \vec{E_0} [/itex] where [itex] \vec{E_0} [/itex] is the field that would be present if both slabs were removed. You cannot use the analysis below Figure 4.21 to show that because this is definitely a case that Griffiths warns us about in the paragraph above Figure 4.21. Therefore I do not see why [itex]\vec{D} [/itex] and [itex] \vec{E_0} [/itex] are parallel.

**2. Homework Equations**

**3. The Attempt at a Solution**

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