Will Latex make a comeback someday?

[nahya]

never mind. sorry :)

 

[FrogPad]

I don't know where my calculator is, so this might be wrong... but the theory should be there.

Qa
Qb
are point charges.

Thus:
\vec E = \frac{kq}{r^2}\hat r

\vec r_{PQa} = \left[\begin{array}{c} 11 \cross 10^2 m \\ 0 \\ 0 \end{array} \right]

\vec r_{PQb} = \left[ \begin{array}{c} (11+22) \times 10^2 m \\ 0 \\ 0 \end{array} \right]

r_{PQa} = |vec r_{PQa}| = \sqrt{(11\times 10^2 m)^2+(0)^2+(0)^2}=11\times 10^2m

r_{PQb}=33\times 10^2 m

\vec E_{net} = \vec E_{Qa} + \vec E_{Qb} = \frac{kQa}{r_{pQa}^2}\hat r_{Qa} + \frac{kQb}{r_{{pQb}^2}\hat r_{Qb}

Notice that:
\hat r_{Qa}=\hat r_{Qb} = \hat i = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]

Thus:
\vec E_{net} = k\hat i \left(\frac{4\times 10^{-6}}{(11 \times 10^2 m)^2} + \frac{-1 \times 10^{-6}}{33 \times 10^2 m}\right)

After you plug in k and cross of the units you get:
\vec E = \hat i \left((9\times 10^9) \frac{4\times 10^{-6}}{(11 \times 10^2 m)^2} + \frac{-1 \times 10^{-6}}{33 \times 10^2 m}\right)\fraq{N}{C}

Also remember that you can drop the vector notation for \vec E_{net} [/tex] by taking the magnitude of it. And since you only have one component E_{net} is just equal to whatever that calculation is above. Sorry, I really don't feel like doing that by hand... not that it's hard. I'm just too lazy to do it.<br /> <br /> For the second part of your question recall that: <br /> \vec F_{12} = \frac{kq_1 q_2}{r^2}\hat r<br /> <br /> I think you can figure it out from what I showed you above. I gave a very rigrious calculation, not skipping many steps. I find it easier just to think about most questions like this as three dimensional questions, and just do a little bit more steps with the vector calculus.

 

[nahya]

well.. the latex thing isn't working atm, i guess. ^^
i solved the problem, however. i just didn't think about the types of the resulting charges.

 

[FrogPad]

Why are you taking the absolute value? How are you setting up the problem for the force?

 

[FrogPad]

Cool. Well I guess you're good.

Maybe some day Latex will come back :)