Question: I solved the first part and I just need help understanding the second part. This will take me a a few min to write and draw....bear with me...

- anonymous

- katieb

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- anonymous

Point charges of \[q_1=-5.00 \mu C\]\[q_2=+3.00 \mu C\]\[q_3=+5.00\mu C\] located on the x access at \[x=-1.00cm\]\[x=0\]\[x=+1.00cm\]respectively. I calculated \[\vec{E}\] on the x axis atx=3.0cm and x=15.0cm
and got:
\[\vec{E}_{p1}=(1.14\times10^8N/C)\hat{i}\]
and
\[\vec{E}_{p1}=(1.74\times10^6N/C)\hat{i}\]
Here is where I'm having trouble:
ARe there any points on the x-axis where the magnitude of the electric field is zero? If so, where are those points?

- anonymous

- anonymous

typo: I meant to write \[\vec{E}_{p2}=(1.74\times10^6N/C)\hat{i}\]

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## More answers

- anonymous

@Luis_Rivera I'll post the answer, can you help me understand it? I have the solution but I'm having a hard time understanding the concept

##### 1 Attachment

- anonymous

- anonymous

|dw:1360443203643:dw|

- anonymous

Think of it this way: if you were just slightly to the left of the particle at x = -1, the electric field would point to the right, correct?

- anonymous

correct

- anonymous

Now go a hundred billion miles to the left. The distances between the three would be almost zero, so you'd really only be able to see a single particle of charge -5 + 3 + 5 = 3. Does that argument make sense?

- anonymous

Yes, so we would see a particle of charge three. Oh so is \[r\rightarrow \infty\]?

- anonymous

No. The point of that is if you're close to the one at -1, then the field points to the right. If you're far away, the field points to the left, so there must be some point at which the field is zero.

- anonymous

I guess that kinda makes sense....where from here?

- anonymous

The fields from the three particles are:
\[E_1 = \frac{-5}{(x+1)^2} \]
\[E_2 = \frac{3}{x^2} \]
\[E_3 = \frac{5}{(x-1)^2} \]
So the equation you should be solving is
\[E_{total} = E_1 + E_2 + E_3 = 0 \]

- anonymous

\[E_1=\frac{q_1}{(x+1)^2}\]
this means:
the electric field due to a point charge from a positive test particle "x" located at a location (which I'm solving for)
\[E_{total} = E_1 + E_2 + E_3 = 0\]
by solving for x, I've found a place on the x axis where the electric field due to all point charges is zero....

- anonymous

makes sense. THanks @Jemurray3 !

- anonymous

The positive and negative charges are bothering me...|dw:1360445236763:dw|

- anonymous

What's bothering you?

- anonymous

I Would've written this as:
\[-\frac{q_1}{(x-1)^2}-\frac{q_2}{x^2}-\frac{q_3}{(x+1)^2}=0\]

- anonymous

because the charge on q_1 is negative

- anonymous

You must also take into account the direction the field is pointing. The field from the q_1 charge is to the right.

- anonymous

Generally speaking
\[ \vec{F} = \frac{kQ}{r^2} \hat{r} \]
So Q is negative, but that is compensated by the direction of r hat.

- anonymous

That explains the positive sign, what about the x_1 part? why do they have it as x-(-1)?

- anonymous

I mean the x-1 part

- anonymous

That is the distance between a point (x) and the point (-1).

- anonymous

which should be x-1 though. That doesn't make sense in practice...I think....
Ok let's see (I can't believe I'm struggling with simple arithmetic)
\[\triangle x=|x_2-x_1|\]
so let's say our point is at x_2=-5
Oh I see....nevermind haha yep it's plus (+)

- anonymous

If you ever forget, just say "okay, so where is the distance zero?" and if it matches the location of the actual particle then you're good.

- anonymous

so (x+1) is zero at x=-1 , which is where the particle is.

- anonymous

yep, I'll remember that. Thanks again @Jemurray3 !!!

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