Columb's Law Questions

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• Oct 3rd 2007, 09:06 AM
Xfd
Columb's Law Questions
Hi,
I am new to the forum as some of you will have noticed and I was quite impressed with what I have seen as lots of useful information can be found here.

Im not very good at Maths, Physics at all but was wondering if any of you could help me please, but only if you dont mind.

H1 FOE.jpg (Click link to view question)

In this question im guessing because the length between particle 1 and 2 is 9.00cm that particle 3 will have to be located above the two points giving an equilaterial triangle.

From this we know all the sides will be the same length of 9.00cm with an angle of 60 degrees. However, there is no solid value given for the charge on particle 1 and 2.

Since
Q1 has a charge of +Q
Q2 has a charge of +4.00Q

So does this mean that the above forumula has to be arranged somehow,
for F3,1 and F3,2 ?

Im not too sure, it would be helpful if someone could tell me what the correct formula is.

Thank you.
• Oct 3rd 2007, 09:39 AM
topsquark
Quote:

Originally Posted by Xfd
Hi,
I am new to the forum as some of you will have noticed and I was quite impressed with what I have seen as lots of useful information can be found here.

Im not very good at Maths, Physics at all but was wondering if any of you could help me please, but only if you dont mind.

H1 FOE.jpg (Click link to view question)

In this question im guessing because the length between particle 1 and 2 is 9.00cm that particle 3 will have to be located above the two points giving an equilaterial triangle.

From this we know all the sides will be the same length of 9.00cm with an angle of 60 degrees. However, there is no solid value given for the charge on particle 1 and 2.

Since
Q1 has a charge of +Q
Q2 has a charge of +4.00Q

So does this mean that the above forumula has to be arranged somehow,
for F3,1 and F3,2 ?

Im not too sure, it would be helpful if someone could tell me what the correct formula is.

Thank you.

If Q3 is to remain in place then the net force on that charge will have to be 0. If you think about this, since forces are vectors, you cannot have Q3 as the third point of a triangle because there will always be a component of the net force directed either toward or away from the line containing the other two charges.

No, Q3 will have to lie somewhere on the same line as Q1 and Q2. Since we don't even know what the sign of the charge of Q3 will be (though logic says it must be negative) I will simply call the position of Q3 as x. (The y coordinate is 0, of course.) I am going to assume for the moment that Q3 is between charges 1 and 2 until I run into a contradiction. So our assumptions are the Q3 is negative and between the other two charges.

So let's calculate the magnitude of the force on Q3 by Q1:
$F_{31} = \frac{kQ_1Q_3}{x^2} = \frac{kqQ_3}{x^2}$
Since this force is toward Q1 the force component is negative:
$F_{31} = - \frac{kqQ_3}{x^2}$

The magnitude of the force on Q3 by Q2 is:
$F_{32} = \frac{kQ_2Q_3}{(9 - x)^2} = \frac{4kqQ_3}{(9 - x)^2}$
Since this force is toward Q2 the force component is positive:
$F_{32} = \frac{4kqQ_3}{(9 - x)^2}$

Now the sum of these forces is required to be 0:
$F_{31} + F_{32} = 0 = - \frac{kqQ_3}{x^2} +
\frac{4kqQ_3}{(9 - x)^2}$

We may solve this equation for x:
$0 = - \frac{kqQ_3}{x^2} +
\frac{4kqQ_3}{(9 - x)^2}$
<-- Cancel out the extra stuff:

$0 = - \frac{1}{x^2} +
\frac{4}{(9 - x)^2}$

$\frac{1}{x^2} =
\frac{4}{(9 - x)^2}$

$(9 - x)^2 = 4x^2$

$x^2 - 18x + 81 = 4x^2$

$3x^2 + 18x - 81 = 0$

$x^2 + 6x - 27 = 0$

Typically Physics equations don't factor, but this one does:
$(x + 9)(x - 3) = 0$

So x = 3 cm or x = -9 cm.

Does the x = -9 cm solution make any sense? No, because at this point both Q1 and Q2 would be pulling it in the +x direction. So we discard this solution.

So let's look at the net force on Q1 (we could also look at Q2):
$F_{12} = \frac{kQ_1Q_2}{9^2} = \frac{4kq^2}{81}$
This force is repulsive (pushing Q1 in the -x direction) so the force component is negative:
$F_{12} = - \frac{4kq^2}{81}$

and
$F_{13} = \frac{kQ_1Q_3}{3^2} = \frac{kqQ_3}{9}$
This force is attractive (pulling Q1 in the +x direction) so the force component is positive:
$F_{13} = \frac{kqQ_3}{9}$

And again the net force on Q1 is 0:
$F_{12} + F_{13} = 0 = - \frac{4kq^2}{81} + \frac{kqQ_3}{9}$

We may solve this equation for Q3/q:
$0 = - \frac{4kq^2}{81} + \frac{kqQ_3}{9}$

$0 = - \frac{4q}{9} + Q_3$

$\frac{4q}{9} = Q_3$

$\frac{Q_3}{q} = \frac{4}{9}$

I'll let you verify that the net force on Q2 is also 0.

-Dan
• Oct 3rd 2007, 09:52 AM
Xfd
Oh I see, I will give that formula a try to see what answer I come out with.

Thanks very much for answering my question.
I appreciate it.
• Oct 7th 2007, 04:19 AM
Xfd
I also have another question that I am very confused about :confused:

I think you are suppose to find the magnitude and direction of the electric field and the formula needed is,

E = 1 / (4x 3.14 x e0) x (density x distance / R^2)

And I was told integration is needed too??

but I really dont know where to start, a correct formula to show the steps would be helpful.

Many thanks !
• Oct 7th 2007, 07:05 AM
topsquark
Quote:

Originally Posted by Xfd
I also have another question that I am very confused about :confused:

I think you are suppose to find the magnitude and direction of the electric field and the formula needed is,

E = 1 / (4x 3.14 x e0) x (density x distance / R^2)

And I was told integration is needed too??

but I really dont know where to start, a correct formula to show the steps would be helpful.

Many thanks !

First, new questions should go in new threads. This is NOT a Coulomb's Law question!!

You are to apply the formula:
$\vec{E} = \frac{1}{4 \pi \epsilon _0} \int \frac{\lambda(\vec{r}^{\prime}) ds^{\prime}}{R^2}$
where the integral is over the line charge, $\vec{R}$ is the displacement from $ds^{\prime}$ (the infinitesimal charge element) to the point where the field is being evaluated.

In this case, $\lambda$ is a constant and is equal to
$\lambda = \frac{\pm 4.50~pC}{\frac{2\pi r}{4}}$
with the + or - depending on if we are in the upper or lower quadrant of the circle.

So let's do each quadrant of the circle separately and then add the two E fields.

The upper quadrant:

One thing to remember when doing this is that the direction of $\vec{R}$ changes for each charge element $ds^{\prime}$ on the line. So what I would do here is break the E field up into x and y components, where I am defining the usual coordinate system. So

$E_{up,x} = \frac{1}{4 \pi \epsilon _0} \int \frac{\lambda ds_x^{\prime}}{r^2}$

Now,
$ds_x^{\prime} = dx^{\prime} cos(\theta ^{\prime} )$ where $dx^{\prime}$ is the x coordinate of $ds^{\prime}$ and $\theta ^{\prime}$ is the angle between $\vec{R}$ and the -x axis.

Now, $cos(\theta ^{\prime} ) = \frac{x^{\prime}}{r}$ where $x^{\prime}$ is the x coordinate of $ds^{\prime}$.

So:
$ds_x^{\prime} = \frac{x^{\prime}}{r} \cdot dx^{\prime}$

Thus
$E_{up,x} = \frac{1}{4 \pi \epsilon _0} \int _0^r \frac{\lambda \frac{x^{\prime}}{r} \cdot dx^{\prime}}{r^2}$

$E_{up,x} = \frac{\lambda}{4 \pi \epsilon _0 r^3} \int_0^r x^{\prime} dx^{\prime}$

$E_{up,x} = \frac{\lambda}{4 \pi \epsilon _0 r^3} \int_0^r x^{\prime} dx^{\prime}$

$E_{up,x} = \frac{\lambda}{8 \pi \epsilon _0 r}$

Now, what direction is the x component of the E field? Since the charge in the upper quarter circle is positive the field must be toward point P. (This interpretation is supported by the overall positive sign of E on the component implying the field is in the +x direction.) Thus
$\vec{E_{up,x}} = \frac{\lambda}{8 \pi \epsilon _0 r} \hat{i}$

Now, you do the $E_{up, y}$ component and the integrals for the lower quarter circle.

Once you get your answer, do you notice anything in retrospect that could have provided a faster solution method?

-Dan
• Oct 7th 2007, 07:31 AM
Xfd
Going back to the first question...

Does this mean that because the charges on particle 1 and 2 (Q1 and Q2) are both positive that they will repel, and the charge on particle 3 will need to be negative to hold the particles in place ?

Also in the formula you had written,

kq x Q3 / x^2

but where does the "q" bit come in from the formula

K x Q1 x Q3 / x^2

???

And Im guessing that the "9 - x" part in the formula is the

Total length between Q1 and Q2 (9.00cm) - subtracted by the distance between Q1 and Q3 (x).

Where does the "3" come from in front of the "x" in the equation?
I understand that afterwards each side is divide by 3 but cant remember why it appears there in the first place.

The signs change and the 4 is gone?
• Oct 7th 2007, 07:43 AM
topsquark
Quote:

Originally Posted by Xfd
Going back to the first question...

Does this mean that because the charges on particle 1 and 2 (Q1 and Q2) are both positive that they will repel, and the charge on particle 3 will need to be negative to hold the particles in place ?

Yes.

Quote:

Originally Posted by Xfd
Also in the formula you had written,

kq x Q3 / x^2

but where does the "q" bit come in from the formula

K x Q1 x Q3 / x^2

???

There is a bit of a notational problem between what you wrote and what the question states. Perhaps I should have been clearer: I'm labeling $Q_1 = +q$ and $Q_2 = +4.00q$. $Q_3$ I just left as $Q_3$.

Quote:

Originally Posted by Xfd
And Im guessing that the "9 - x" part in the formula is the

Total length between Q1 and Q2 (9.00cm) - subtracted by the distance between Q1 and Q3 (x).

I'm putting the third charge at a position x between Q1 and Q2. So the distance between Q1 and Q3 is x, and the distance between Q2 and Q3 is 9 - x.

Quote:

Originally Posted by Xfd
Where does the "3" come from in front of the "x" in the equation?
I understand that afterwards each side is divide by 3 but cant remember why it appears there in the first place.

The signs change and the 4 is gone?

$x^2 - 18x + 81 = 4x^2$

Move everything to one side of the equation:
$4x^2 - (x^2 -18x + 81) = 0$

$3x^2 + 18x - 81 = 0$

-Dan
• Oct 7th 2007, 08:12 AM
Xfd
Quote:

Originally Posted by topsquark
$x^2 - 18x + 81 = 4x^2$

Move everything to one side of the equation:
$4x^2 - (x^2 -18x + 81) = 0$

$3x^2 + 18x - 81 = 0$

-Dan

I see now because the minus "-" sign changes the signs when multiplying out the brackets and $4x^2 - x^2 = 3x^2$

Quote:

Originally Posted by topsquark
I'm putting the third charge at a position x between Q1 and Q2. So the distance between Q1 and Q3 is x, and the distance between Q2 and Q3 is 9 - x.

So that means you would have something like this.

http://i122.photobucket.com/albums/o...s5/diagram.jpg
• Oct 7th 2007, 08:14 AM
topsquark
Quote:

Originally Posted by Xfd
I see now because the minus "-" sign changes the signs when multiplying out the brackets and $4x^2 - x^2 = 3x^2$

So that means you would have something like this.

http://i122.photobucket.com/albums/o...s5/diagram.jpg

Yes to both. :)

-Dan
• Oct 7th 2007, 08:23 AM
Xfd
Okay thanks for your help :)
That looks like everything I need then for the first question.

I will look over what you wrote for the second question to see If I understand first.
• Oct 7th 2007, 10:37 AM
Xfd
Quote:

Originally Posted by topsquark
Thus
$\vec{E_{up,x}} = \frac{\lambda}{8 \pi \epsilon _0 r} \hat{i}$

Now, you do the $E_{up, y}$ component and the integrals for the lower quarter circle.

Once you get your answer, do you notice anything in retrospect that could have provided a faster solution method?

-Dan

Okay so you are saying afterwards you have to find the Y component for the upper quadrant with the formula:

$\vec{E_{up,y}} = \frac{\lambda}{8 \pi \epsilon _0 r} \hat{i}$

And then the lower quadrant with the equations:

$\vec{E_{down,x}} = \frac{\lambda}{8 \pi \epsilon _0 r} \hat{i}$

$\vec{E_{down,y}} = \frac{\lambda}{8 \pi \epsilon _0 r} \hat{i}$

???