User:Frumfst

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You have new messages (last change).

As quoted by Kurt Vonnegut in Slaughterhouse-Five,

"That was I. That was me. That was the author of this book."

Meaning, don't mess with me, becuase I do not like it when other people put down false information. I won't.

Since I have nothing important to say, I'll just make sections of my favorite whatevers.

Contents

[edit] Favorite Equation

\vec{F} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{|\vec{r}|^2} \hat{r}

For those that don't know

\varepsilon_0 = \frac{1}{c^2 \mu_0} \approx 8.8541878176 × 10-12 F/m (C2/Jm)

By subsistution, it is possible to get the equation to be:

\vec{F} = \frac{1}{4 \pi \frac{1}{c^2 \mu_0}} \frac{q_1 q_2}{|\vec{r}|^2} \hat{r} which, by simplifying, becomes \vec{F} = \frac{c^2 \mu_0}{4 \pi} \frac{q_1 q_2}{|\vec{r}|^2} \hat{r}

If you really want to you can take \frac{1}{4 \pi \epsilon_0} or \frac{1}{4 \pi \frac{1}{c^2 \mu_0}} or \frac{c^2 \mu_0}{4 \pi} and change it to the electrostatic constant kC to get:

\vec{F} = k_C \frac{q_1 q_2}{|\vec{r}|^2} \hat{r}

Now, if you are really lucky and q1 = q2, the equation becomes:

\vec{F} = k_C \frac{q^2}{|\vec{r}|^2} \hat{r}

Let's say that the two charges are equal and both are on the x-axis, then:

\vec{F} = \pm k_C \frac{q^2}{r^2} and the directions will always being in opposite direction, so there's no need for the unit vector. The plus-minus sign just means one charge will go to the right while the other will go to the left.

However, the only time you will ever use the uber simplified equation is when you are in baby physics.

[edit] Favorite Calculator

Marvel at its superiority
Marvel at its superiority

The TI-89 Titanium (as pictured)

[edit] Favorite Time Zone

Pacific

[edit] LOOK AT THIS

This user lives in the state of Washington.
This user plays the piano.


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