Brief Physics Intercision – Force due to Electric charges between varying dielectrics

Allow me to interrupt the (irregular) schedule of the Gospel with a Physics question.

I study Electrical and Electronic Engineering in Lausanne, Switzerland. In a recent Electrotechnique I class, the topic of force due to an electric charge came up. Basically, for two positive electric charges Q_1 and Q_2 in a vacuum (permitivity \epsilon_0, a distance r), the force due to the electric charges can be expressed as the following:

F_1 = F_2 = F = \frac{Q_1 Q_2}{4 \pi \epsilon_0 r^2}

However, that supposes that the dielectric between them, in this case a vacuum, is the same. What I asked today in class, and this was something the professor was incapable of answering on the spot, was: how do you express the force if Q_1 is in one dielectric \epsilon_1 and Q_2 is in another dielectric \epsilon_2? Let’s say for convenience, there is a clean separation of the two dielectrics at a point, say, \frac{r}{2}.

Any help is greatly appreciated.

Layman’s terms:

Given two ideal, solitary, positive electric charges, a theme very commonly used to explain electromagnetism equations in Physics, you can express the force they exert upon each other. However, it assumes that the material which lets through the electric field caused by each other is the same throughout the system. What if one charge was in one type of ‘material’ and the other in another type? It’s a simple enough concept to imagine, and it’s very easily answerable if you’re talking about capacitors (a type of electronic component that stores electric charge).

Thanks in advance.


3 Responses

  1. Would you happen to know how I would post some equations in this space, or in reddit where I have responded similarly?

    Basically the procedure is to find D, and thus P, and thus E at the location of the second charge, if I’m reading this textbook correctly. A bit rusty, I am, but I’ve found the right chapter.

  2. Thinking about it, this may work out to simply a factor of epsilon_1/epsilon_2 on the original equation

  3. Actually, a factor of ep_1/ep_2 plus the polarization attraction between the test charge and the junction.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: