Question

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...

Answer 1

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?

Answer 2

@Luis_Rivera @phi

Answer 3

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

Answer 4

@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

Answer 5

@Jemurray3

Answer 6

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?

Answer 7

correct

Answer 8

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?

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

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
\[E_{total} = E_1 + E_2 + E_3 = 0 \]

Answer 13

\[E_1=\frac{q_1}{(x+1)^2}\]
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....

Answer 14

makes sense. THanks @Jemurray3 !

Answer 15

The positive and negative charges are bothering me...

Answer 16

What's bothering you?

Answer 17

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

Answer 18

because the charge on q_1 is negative

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

I mean the x-1 part

Answer 23

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

Answer 24

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 (+)

Answer 25

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.

Answer 26

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

Answer 27

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