Recent content by jmtome2

Homework Statement Show graphically how \vec{a}\times\vec{x}=\vec{d} defines a line. \vec{a} and \vec{d} are constants. \vec{x} is a point on the line.Homework Equations \vec{a}\times\vec{x}=a\cdot x\cdot sin(\theta)\cdot \hat{n}The Attempt at a Solution Not sure if the included relevant...

YES! Of course it is... what was I thinking. But this creates a whole problem... I've got to figure out how to plot E now inside the program without using the ODEsolver, which hasn't been mentioned in the book yet. *sigh* Thanks for clarifying :)

Homework Statement OK so here goes. I'm using an ODEsolver in java to plot the total energy over time of a planetary system. So I've been trying to calculate the rate of energy (per unit mass), \frac{E}{m}. Homework Equations The equation for total energy (per unit mass) of a planetary...

I know the capacitance of a parallel plate configuration with dielectric filling the middle is C=\frac{A\epsilon_0\epsilon_r}{d}, but I also know that the grounded "plate" on top throws this off... I'm just not sure how

b]1. Homework Statement [/b] Electrostatic clamps are used for holding workpieces while they are being machined, for holding silicon wafers during electron beam microfabrication, etc. They comprise an insulated conducting plate maintained at a potential of several thousand volts and covered...

Haven't really dealt with grounded conductors before... this means that extra charge rolls off of these plates leaving them with Q=0, meaning that they don't contribute to the E field?

Homework Statement "When a block of insulating material such as Lucite is bombarded with high-energy electrons, the electrons penetrate into the material and remain trapped inside. In one particular instance a 0.1 mA beam bombarded an area of 25cm^{2} of Lucite (Class A: \epsilon_r=3.2) for 1...

That's it! I always miss the small things, thanks

sounds good to me, I'm actually asked later to calculate the net charge per unit length so I'll need that. And since we are on the topic... :) to get the net charge per unit length at r=a, \frac{Q_{net}}{L}=\frac{Q_{free}}{L}+\frac{Q_{bound}}{L}? I'm told that the answer is...

urgh... so should I assume that its finite or assume that it is infinite? lol... is there some way of calculating the inner and outer surface bound charge density so that it doesn't matter?

Doesn't say that it is so I just assumed it was length L

Quick final question: For the bound charge (on both the inner and outer surface) final answer I get Q_{bound}=\frac{\epsilon_0\cdot \chi_{e}}{\epsilon}\cdot \lambda L but doesn't \lambda L=Q_{free} , where Q_{free} is the free charge of the configuration?

\textbf{E}_{\text{slab}} is the electric field from the whole chunk of dielectric??

Cool and thanks for your help

Another valid point \vec{D}=\frac{\lambda}{2\pi r} \cdot \hat{r} and therefore, \vec{E}=\frac{\lambda}{2\pi \epsilon\cdot r} \cdot \hat{r}