• Support PF! Buy your school textbooks, materials and every day products Here!

The missing variable

  • Thread starter Physicsit
  • Start date
  • #1
Physicsit
Ok this seems easy enough but I am missing something

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.200 m. In a vacuum, each object carries a different charge, and they at-tract each other with a force of 1.20 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the ob-jects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object?

Coulumbs law states

that

f=k * q1*q2/r^2

f is clearly given
r is clearly given

k is a constant 8.99*10^9

1.20=8.99*10^9 * q1*q2/.200^2

where do q1 and q2 come into the picture?
 

Answers and Replies

  • #2
AD
72
0
Well, it says that the charge is shared equally between the two objects, so just call the charge on each object q.

Then

F = k ((q * q)/r2) = (kq2)/r2

Now re-arrgange to find q:

q2 = (Fr2)/k

and

q = ±√[(Fr2)/k]

Since it didn't state whether the charge was positive or negative.

I'll leave you to compute the answer.
 
  • #3
AD
72
0
Whoops. It appears that I didn't actually solve the problem. But I think you can get the two charges' sum and product from this and you might be able to find them that way.
 
  • #4
Doc Al
Mentor
44,940
1,201
Originally posted by AD
Whoops. It appears that I didn't actually solve the problem. But I think you can get the two charges' sum and product from this and you might be able to find them that way.
Right. Call the final charge on each q; use Coulomb's law to find q.

If the original charges are q1 & q2, you know what the must add up to (2q) and what their product must be (-q2). Do a little algebra and you'll figure them out.
 
  • #5
Physicsit
So does this look correct

So does this seem correct

f=k(q^2/r^2)

q = +/-squareroot([(Fr2)/k])

q=+/-squareroot([(1.20*(.200^2))/(8.99*10^9)])

q=+/- 2.31*10^-6


it would appear so because




8.99*10^9*(2.31*10^-6*2.31*10^-6)/.200^2 =1.20N

so Is this the right way to do it?
 
  • #6
1,036
1
Not quite.
q=+/- 2.31*10^-6
This is true, but you're not finished. (Better make it +/- 2.311*10-6).
Not only does
q1 ≠ q2
but
q1 ≠ -q2

Do you see why?

If
q1 = -q2
then after the objects touch the final charge q would be 0 and there would be no electrostatic force between them.
 
  • #7
Physicsit
Kind of lost then

I am not sure where to go with it then


I would have thought q1 = -q2

I am not sure how to solve for q1 and q2



with q=+/- 2.311*10^-6

it says that the objects are identical and initialy they repel each other so they have to be opposites
so I am not sure

why q1 cannnot = - q2

it would seem that how it should be since

they objects are identical except for the charge
 
  • #8
Doc Al
Mentor
44,940
1,201


Originally posted by Physicsit
I would have thought q1 = -q2
I thought gnome explained this point well. When the two objects touch, they will distribute the charge between them equally. We know there must be a non-zero charge (q) on each object after they touch since they repel. So q1 cannot equal -q2: since, if they did, there would be no charge on either object after they touch. Do not continue until you understand this point.
it says that the objects are identical and initialy they repel each other so they have to be opposites
The objects are identical, but not their charges. All you can deduce from the fact that they initially attract is that the two charges (q1, q2) are opposite in sign. You cannot conclude that the magnitudes are equal.
so I am not sure

why q1 cannnot = - q2

it would seem that how it should be since

they objects are identical except for the charge
Read the above carefully.

To solve for q1 & q2:
First find q (the final charge on both objects).

Then consider these two conditions that q1 and q2 must satisfy:
q1 + q2 = 2q (this comes from the fact that the total charge will distribute equally after touching)
q1 x q2 = -q2 (this comes from the fact that the force is equal, but opposite, before and after touching; look at Coulomb's law)

Combine these two equations and solve for q1 & q2. But don't waste time solving until you understand where these two equations come from and what they mean.
 
  • #9
1,036
1
Yes, the objects are identical except for the charge.
And since they ATTRACT each other, one must be positive and the other must be negative.
So let one initial charge be X and the other one Y. One of these numbers is negative, but you'll worry about that later. Meanwhile just call them X and Y.
Since they are identical except for the charge, if one has charge X and the other has charge Y, after they touch, the final charge q on EACH of them will be (X+Y)/2.
But if X = -Y, then (X+Y) = 0
and then each one would have 0 charge and there would be no force between them.

So, you have two equations and two unknowns. Messy equations, because of the numbers you're dealing with, but its still just algebra at this point. You already solved for q, and you know that
X + Y = 2q
so that's one equation.
The other one is your original equation:
KXY/r2 = 1.2

You'll end up with a messy quadratic equation to solve.

Sorry, I switched to X and Y to avoid typing all those subscripts. Hope that didn't confuse you.
 
  • #10
1,036
1
Note that there is no contradiction between Doc Al's second equation and mine.

Remember, way back when you solved for q, you said:
q2 = (1.2 * r2)/k

If you rearrange my second equation slightly you get:
XY = (1.2 * r2)/k

and I said that either X or Y is negative.

So XY = -q2
 
  • #11
Physicsit
I think I understand

Thanks for all the help I think I got it now

We know that

q1 + q2 = 2q

and

q1 x q2 = q ^2

we found q earlier 2.311*10^-6

so

q^2 = 5.34*10^-12
2q = 4.66*10^-6

so

5.34*10^-12=q1 * q2

4.66*10^-6 = q1 +q2

so

4.66*10^-6 - q1 =q2

so

5.34*10^-12=q1 * (4.66*10^-6 - q1)

0=-q1^2 + 4.622*10^-6 - 5.34*10^-12

plug it all into the quadratic equation

q1 = - 4.622*10^-6 +/- squareroot(((4.622*10^-6)^2 -(4)(-1)(5.34*10^-12))/(2(-1))

q1= - 4.622*10^-6 + 3.27*10^-6

q1 = -1.352*10^-6
so
4.66*10^-6 - -1.352*10^-6 =q2
q2=5.974*10^-6

or

q1= - 4.622*10^-6 - 3.27*10^-6

q1= -7.89 * 10^-6

so

4.66*10^-6 - -7.89 * 10^-6 =q2
q2 =12.514*10^-6
 
  • #12
1,036
1
Nope.

After you solve, you have to check your answer & see if it makes sense. At least, try multiplying q1 x q2 & see what you get.

Here's the problem:
We know that

q1 + q2 = 2q

and

q1 x q2 = q ^2
No! q1 x q2 = -q^2
That little minus sign makes a big difference.

As a result, this:
0=-q1^2 + 4.622*10^-6 - 5.34*10^-12
should be:
0=q12 - 4.622*10-6 - 5.34*10-12
 

Related Threads on The missing variable

  • Last Post
Replies
4
Views
4K
Replies
10
Views
9K
Replies
3
Views
1K
  • Last Post
Replies
3
Views
918
  • Last Post
Replies
4
Views
4K
Replies
7
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
0
Views
1K
Top