Calculate Net Force on Q3: Charge Position for Zero Net Force

[Winzer]

Homework Statement

The charges Q1= 1.90·10-6 C and Q2= -3.03·10-6 C are fixed at their positions, distance 0.279 m apart, and the charge Q3= 3.33·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.

Homework Equations

F=\frac{KQq_{1}q_{2}}{r^2}

The Attempt at a Solution

I get 0.6483m

 

[Doc Al]

Please give the coordinate of each particle along an axis.

 

[rootX]

is the answer 1.06?

 

[Winzer]

Q(0,0), Q2(.279,0), Q3(x,0). Where X>.279m

 

[Doc Al]

Q(0,0), Q2(.279,0), Q3(x,0). Where X>.279m

That's what I thought you were saying. Note that X > .279 is not a requirement of the problem, but something that you added. In fact, it can't be true!

 

[Winzer]

True, i did add that in.
I started by finding the force between q1 and q2:
F=\frac{9.0e^9 1.90e^-6 -3.03e^-6}{.279^2}=-6.65e^-1N
Then I can set that equal to F=\frac{KQq_{1}q_{2}}{(x-r)^2}, right?

 

[rootX]

just making it way too complex,
find where E is 0, and that would be the position of Q3
and you would get something like
q/r^2 = q1/(d+r)^2

 

[Doc Al]

I started by finding the force between q1 and q2

The force between Q1 and Q2 is irrelevant. You want the force on Q3! (Hint: Where does the electric field of Q1 + Q2 equal zero?)

The first thing I would do is figure out which region Q3 must be in:
(a) to the left of Q1
(b) between Q1 and Q2
(c) to the right of Q2

 

[Winzer]

So: E_{1}+E_{2}=0
\frac{kq_{1}}{x^2}-\frac{kq_{2}}{(r+x)^2}=0

 

[Doc Al]

So: E_{1}+E_{2}=0
\frac{kq_{1}}{x^2}-\frac{kq_{2}}{(r+x)^2}=0

What region are you analyzing? What's the definition of x?

If it's not obvious which region Q3 must be in, study each region systematically.

 

[Winzer]

the region right of q2. X is distance from q1 to q3

 

[Doc Al]

the region right of q2. X is distance from q1 to q3

For some reason you are choosing the region to the right of q2 (even though I told you that it was impossible! ). No problem! If x is the distance from q1, what must be the distance from q2 in that region?

 

[Winzer]

ok, so I the region where E=0, q3 would have to inbetween q1 & q2. getting confused

 

[Doc Al]

ok, so I the region where E=0, q3 would have to inbetween q1 & q2. getting confused

Don't give up on studying any region. If you pick the wrong region, the equation you get will not give you an answer in the region.

To study the region in between q1 & q2, draw yourself a picture. In that region, which way would the field from q1 point? The field from q2? Could they cancel?

 

[Winzer]

I know the q1 field is outward and q2 field in inward. So the field(right side) from q1 to q2 will interect with each other.

 

[Doc Al]

I know the q1 field is outward and q2 field in inward.

Right! So, in the region between q1 and q2, which way does each field point: to the right or to the left? If we let positive stand for "to the right", are the fields positive or negative in that region?

 

[Winzer]

positve to the right, negitive to the left

 

[Doc Al]

positve to the right, negitive to the left

That's the sign convention. But which way do the fields point?

 

[Winzer]

The combination of the q1 and q2 field points to the right.

 

[Doc Al]

The combination of the q1 and q2 field points to the right.

In that middle region, the field from q1 points to the right (and thus is positive) and the field from q2 also points to the right (and is positive). So is it possible for the fields to cancel out in that region?

 

[Winzer]

mmm..yes

 

[Doc Al]

mmm..yes

Answer this: If A is positive and B is positive, can A+B ever equal zero?

 

[Winzer]

If both A and B are positive than there is an efield that is zero at some point. Because both field lines from A and B will repel each other.

 

[Doc Al]

If both A and B are positive than there is an efield that is zero at some point.

No. If A and B are positive, their sum must also be positive. They can never add to zero. You can only get 2 numbers to sum to zero if they have different signs.

Because both field lines from A and B will repel each other.

The fields from each charge (q1 & q2) point in the same direction and thus add together--they don't cancel. That tells you that the force on q3 will never be zero if it's placed between q1 and q2. (Now you need to check the other two regions.)

 

[Winzer]

so I suspect that the only way the fields repel if we put it before q1?

 

[rootX]

yep, seems right to me

 

[Winzer]

so how do i find where the E field will be zero? Super position of E?

 

[rootX]

yes, you had that formula E1+E2=0

 

[Winzer]

so:
\frac{kq_{3}}{x^2}+\frac{kq_{1}}{(r+x)^2}=0
now i don't uderstand the (r+x)^2 part.

 

[rootX]

so:
\frac{kq_{2}}{x^2}+\frac{kq_{1}}{(r+x)^2}=0
now i don't uderstand the (r+x)^2 part.

so, the E is 0 @ x distance left from the smaller positive charge, and thus we know distance between them is r..

 

[Winzer]

I assume you mean the distance from q3 to q2. r+x. Is that what you meant?

 

[rootX]

nopes,
but you are/ should be using q1 and q2?

 

[Winzer]

thought i am using q3 & q1

 

[rootX]

oh..that was wrong, I guess

 

[Winzer]

so what am I using?

 

[rootX]

q1, and q2

you are simply lost!, don't you? ^^

 

[Doc Al]

It seems that you are still a bit confused. Remember that you are trying to find out where to put Q3 so that it feels no net force from Q1 + Q2. Since Q3 feels the electric field from Q1 + Q2, this is equivalent to asking where is the net field from Q1 + Q2 equal to zero.

Let's look at the field to the left of Q1 at some point a distance X to the left of Q1. The field from Q1 at that point will be:
E_{q_1} = - \frac{k q_1}{x^2}

Note that I put a negative sign, since the field points to the left.

Now the field from Q2 (realize that if the distance from the point to Q1 is X, then its distance from Q2 must be X + 0.279 m):
E_{q_2} = + \frac{k q_2}{(x + 0.279)^2}

Note that this field points to the right, so it's positive.

Now add them up and solve for the distance (x) that makes the sum equal to zero.

Note that in both of the equations above, I let q_1 and q_2 stand for just the magnitude of the charges. (Don't put in a negative sign twice!)

 

[PFStudent]

The charges Q1= 1.90·10-6 C and Q2= -3.03·10-6 C are fixed at their positions, distance 0.279 m apart, and the charge Q3= 3.33·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.

Hey,

In solving this problem refer to the following principle,

---------------------------------------------------------------------------------
Given any two arbitrary un-like sign charges: q_{1} and q_{2}, placed on an axis a distance L from each other. Then, the placement (on that axis) of a charge q_{3} such that the net force on q_{3} due to: q_{1} and q_{2}, will be zero. Can be given as follows,

q_{1}q_{2} < 0 \therefore q_{1}q_{2} \equiv -

<br /> |q_{1}| &lt; |q_{2}|, |\vec{r}_{31}| &lt; |\vec{r}_{32}|, |\vec{r}_{32}| &gt; L<br />

<br /> |q_{1}| = |q_{2}|<br />, No equilibrium exists on that axis.

<br /> |q_{1}| &gt; |q_{2}|, |\vec{r}_{31}| &gt; |\vec{r}_{32}|, |\vec{r}_{32}| &gt; L<br />
---------------------------------------------------------------------------------

I wrote up this principle up a while back, because these type of (electric charge) physics problems come up so often.

Thanks,

-PFStudent