# Problem Understanding Permittivity

Discussion in 'Physics' started by djwinger, Jan 11, 2013.

1. ### djwinger Thread Starter New Member

Jan 3, 2012
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So far I have discovered the permittivity of a material relates to the resistance that occurs in an electric field. The higher the permittivity of a material, the less of an electric field created.

How do I interpret this? Does a smaller electric field mean there is less chance for electron and hole movement (i.e. less electric activity), or does it mean a smaller energy gap (i.e. better current flow)?

Apr 5, 2008
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3. ### djwinger Thread Starter New Member

Jan 3, 2012
24
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Yeah and I could barely understand a word of it. I'm not doing physics. Permittivity has come up in my solid state module though. It was never explained...just values given for silicon and germanium. I'm just trying better to understand what it is.

4. ### WBahn Moderator

Mar 31, 2012
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Imagine a room with two metal plates covering opposite walls. You have a capacitor. The capacitance, C, of the room is the ratio of the amount of (differential) charge, Q, deposited on the walls to the voltage, V, that results, namely C=Q/V or Q=CV. The voltage is the electric field intensity, E, (which, initially, we are assuming is uniform) mutiplied by the distance, d, between the walls. The electric field, due to the charge Q, is proportion to the charge density on the walls, for Q/A. For this discussion, we are going to keep most everything fixed, so Q, A, d don't change.

But what if we put some stuff in the room? In particular, imagine we take a bunch of balloons that, while not having a net charge, have positive charges on one side and negative charges on the other. Effectively, each one of these balloons is a charged capacitor and there is a electric field within the balloon and their is a voltage difference from one side of the balloon to the other. While those balloons won't have a tendency to move toward one wall or the other (remember, they have no net charge), they will tend to rotate so that the side that is positively charged will be facing the negatively charged wall and the negatively charged side will be facing the positively charged wall. What happens to the electric field within the balloon? It goes down because it is the sum of the electric field that existed in the room before we put in the balloons and the electric field within the balloon and the balloons oriented themselves so that these fields are in opposite directions. With the average electric field in the room lowered (and the more balloons the more the average field is lowered), the voltage from one wall to the other is lowered. With a lower voltage, the capacitance of the room has increased, meaning that we can put more charge on our room capacitor for the same voltage.

Within many (nearly all, at least to some degree) materials, the atoms/molecules behave as little polarized baloons serving to lower the voltage developed across the material in response to an externally applies electric field (be it from a static charge distribution or otherwise).

Hopefully that will help you develop a useful mental picture of what is going on.

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