# Homework Help: Electric Fields

1. Apr 3, 2006

### nahya

never mind. sorry :)

Last edited: Apr 4, 2006
2. Apr 4, 2006

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 [itex] \vec E_{net} [/tex] by taking the magnitude of it. And since you only have one component [itex] 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.

For the second part of your question recall that:
$$\vec F_{12} = \frac{kq_1 q_2}{r^2}\hat r$$

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.

3. Apr 4, 2006

### 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.

4. Apr 4, 2006