Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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...

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@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
|dw:1360443203643:dw|
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?
correct
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?
Yes, so we would see a particle of charge three. Oh so is \[r\rightarrow \infty\]?
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.
I guess that kinda makes sense....where from here?
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 \[E_{total} = E_1 + E_2 + E_3 = 0 \]
\[E_1=\frac{q_1}{(x+1)^2}\] 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....
makes sense. THanks @Jemurray3 !
The positive and negative charges are bothering me...|dw:1360445236763:dw|
What's bothering you?
I Would've written this as: \[-\frac{q_1}{(x-1)^2}-\frac{q_2}{x^2}-\frac{q_3}{(x+1)^2}=0\]
because the charge on q_1 is negative
You must also take into account the direction the field is pointing. The field from the q_1 charge is to the right.
Generally speaking \[ \vec{F} = \frac{kQ}{r^2} \hat{r} \] So Q is negative, but that is compensated by the direction of r hat.
That explains the positive sign, what about the x_1 part? why do they have it as x-(-1)?
I mean the x-1 part
That is the distance between a point (x) and the point (-1).
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 (+)
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.
so (x+1) is zero at x=-1 , which is where the particle is.
yep, I'll remember that. Thanks again @Jemurray3 !!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question