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JenniferSmart1
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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...
 one year ago
 one year ago
JenniferSmart1 Group Title
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...
 one year ago
 one year ago

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JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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 xaxis where the magnitude of the electric field is zero? If so, where are those points?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@Luis_Rivera @phi
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
typo: I meant to write \[\vec{E}_{p2}=(1.74\times10^6N/C)\hat{i}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@Jemurray3
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
dw:1360443203643:dw
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
correct
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Yes, so we would see a particle of charge three. Oh so is \[r\rightarrow \infty\]?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I guess that kinda makes sense....where from here?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
The fields from the three particles are: \[E_1 = \frac{5}{(x+1)^2} \] \[E_2 = \frac{3}{x^2} \] \[E_3 = \frac{5}{(x1)^2} \] So the equation you should be solving is \[E_{total} = E_1 + E_2 + E_3 = 0 \]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
\[E_1=\frac{q_1}{(x+1)^2}\] 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....
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
makes sense. THanks @Jemurray3 !
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
The positive and negative charges are bothering me...dw:1360445236763:dw
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
What's bothering you?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I Would've written this as: \[\frac{q_1}{(x1)^2}\frac{q_2}{x^2}\frac{q_3}{(x+1)^2}=0\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
because the charge on q_1 is negative
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
You must also take into account the direction the field is pointing. The field from the q_1 charge is to the right.
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
Generally speaking \[ \vec{F} = \frac{kQ}{r^2} \hat{r} \] So Q is negative, but that is compensated by the direction of r hat.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
That explains the positive sign, what about the x_1 part? why do they have it as x(1)?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I mean the x1 part
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
That is the distance between a point (x) and the point (1).
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
which should be x1 though. That doesn't make sense in practice...I think.... Ok let's see (I can't believe I'm struggling with simple arithmetic) \[\triangle x=x_2x_1\] so let's say our point is at x_2=5 Oh I see....nevermind haha yep it's plus (+)
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.2
so (x+1) is zero at x=1 , which is where the particle is.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
yep, I'll remember that. Thanks again @Jemurray3 !!!
 one year ago
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