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JenniferSmart1
 3 years ago
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...
JenniferSmart1
 3 years ago
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...

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JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0Point 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?

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0@Luis_Rivera @phi

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

JenniferSmart1
 3 years ago
Best 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

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1360443203643:dw

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2Think 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?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2Now 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?

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

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2No. 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.

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0I guess that kinda makes sense....where from here?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2The 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 \[E_{total} = E_1 + E_2 + E_3 = 0 \]

JenniferSmart1
 3 years ago
Best 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....

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0makes sense. THanks @Jemurray3 !

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0The positive and negative charges are bothering me...dw:1360445236763:dw

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2What's bothering you?

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

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0because the charge on q_1 is negative

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

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

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

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0I mean the x1 part

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2That is the distance between a point (x) and the point (1).

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0which 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 (+)

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2If 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.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2so (x+1) is zero at x=1 , which is where the particle is.

JenniferSmart1
 3 years ago
Best ResponseYou've already chosen the best response.0yep, I'll remember that. Thanks again @Jemurray3 !!!
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