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Old Feb 17th 2010, 06:20 PM   #1
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electrical charges and forces

hello i am stuck on a problem and was wondering if anyone here can help.

Four identical charges, +Q, occupy the corners of a square with sides of length "a". A fifth charge, q, can be placed at any desired location.
Find the value of q, such that the net electric force acting on each of the original four charges, +Q, is zero.

so i figured that in order for this to be stable q should be negative and in the center of the square. I tried to apply coulombs law F= kq1q2/r^2 to the problem, but im not getting the correct answer.
My answer was q = 2a^2/(4kQ) but the correct answer is:
= . Im not sure how to get this. Im also confused how as to how "a" is not part of the answer. please help
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Old Feb 19th 2010, 05:21 AM   #2
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You have to look at it from the point of one of the outer positive charges. They each have the force from the other three trying to push it away, so you are correct in choosing a negative charge centered between all of them such that the pull from the negative charge cancels out with the push from the positive charges. I suggest that instead of summing the forces on a positive charge and setting those forces equal to zero, since all the forces acting upon a positive charge come from an E field that the charge feels, that we sum the E fields seen by a positive charge and set that equal to zero. No E field seen by the charge means no force acting upon that charge from F = Eq.

Let's start by assigning our unit vectors. i will be for the x axis and j for the y axis. Let's start with the bottom left hand positive charge, set at the origin, and sum the E fields seen to zero. We can find the amount of negative charge needed in the center such that the sum of the E fields in the x direction will be zero. That same amount of charge will keep the sum of all the E fields in the y direction seen by the positive charges equal to zero also, because the value we find is that needed to keep one from moving under the influence of the other three positive charges, so doing one accounts for all three; in other words the E field emitted by the negative charge upon the positive ones will cancel the net E field seen by any of the positive charges.

Let's start by summing the fields acting in the x direction and setting them equal to zero. The equation for the y directed fields will be exactly the same except the unit vectors will be j and not i because the negative charge is centered and acts on all positive charges in the x and y direction with the same amount of force, so once we have the x directed fields, we also have the y directed fields, summing the two will give the total field acting upon a positive charge; as once we find the charge needed to keep a positive charge having a net x directed sum of fields of zero acting on it, so will be the y directed fields. Again typically it would just be the sum of the x and y directed fields, and it is, the sum of something plus something of the same value is simply twice the value of the first something, e.g. (3+3=6 or 2*3 =6). Of course the charge found from summing the x directed fields to zero will be the same charge that will keep the y directed fields to sum to zero as well, so we don’t even need to sum the x and y directed fields and solve for the charge, we can find the charge needed from either the y directed field sum equation or the x directed field sum equation.

OK, the x directed fields.

Fields in x direction = 0 = kq/a^2i + kq/[2sqrt(2)a^2]i + 2kQ/[sqrt(2)a^2]i


Here it is charge by charge, so for the bottom right hand side charge it is just:

E(x) = kq/a^2i and q is the +Q, a is the distance and k = 1/(4*pi*permittivity of the space)

The top right hand side positive charge is sqrt(2)a distance away and this squared should just leave:

E = kq/2a^2 in the direction of a 45 degree angle, and we want just the x directed portion, so we need to multiply this value by 1/sqrt(2) to get just the x directed portion (same thing as cos(45)):

E(x) = kq/[2sqrt(2)a^2]i, and the negative charge in the middle is also off angle so we will have to multiply that field by 1/sqrt(2) once we find it.

It is a/sqrt(2) distance away, so this squared should just be a^2/2, so the field from the negative charge in the middle is:

E = 2kQ/a^2, again we want the x directed portion so let's multiply this by the cos(45) = 1/sqrt(2) we get:

E(x) = 2kQ/sqrt(2)a^2i alright let's sum them:

E(x) = 0 = kq/a^2i + kq/[2sqrt(2)a^2]i + 2kQ/sqrt(2)a^2i

And this is what we got before (it should be), so let’s solve for the charge Q and remember this is the amount of charge that will cause the net fields in the x direction to be zero, but we completely ignored the force from the charges in the y direction, not that it matters because the amount of charge that sums the x directed fields to be zero will also sum the y directed fields to be zero also (because the equation for the sum of the fields in the y direction is the same as in the x direction except the unit vectors are j and not i). Again, usually we would sum the x and y directed fields to be zero, but since this negative charge is right in the middle of four charges in four corners, the fields from the x direction must be the same as the fields in the y direction. Alright let's do some summing and cancelling, (dropping the unit vectors since this equation is the fields in the x direction, knowing that we don't need them):

E(x) = 0 = kq/a^2 + kq/[2sqrt(2)a^2] + 2kQ/sqrt(2)a^2

multiplying each side by a^2 and dividing each side by k yields:

0 = q + q/2sqrt(2) + 2Q/sqrt(2)

now let’s solve for Q and multiply each side of the equation by 2sqrt(2)
-q2sqrt(2) -q = 4Q

Q(x) = -1/4*(q + q2sqrt(2))

and again from E(y) we will get Q(y) = -1/4*(q + sqrt(2)), so we can ignore the y directed fields as they will cancel out just as much as the x directed fields.

So Q(x) = Q(y) = charge needed to sum the fields seen at the positive charges to be zero.

So replace q with Q because the question defines the positive charges to be Q and the charge to be placed which is our negative charge, as q, so we are simply reversing variables, not the values or the meaning of the variables, just stating a is now b and b is now a, so to speak, which is fine, I should have just started by writing the positive charge terms with Q and not q, and the last negative term (not obvious in the equation as it is left positive knowing the signs will work themselves out afterwards) with q instead of Q, but I didn't think of it and again I'm not changing anything really I'm just calling one thing by a different name and another by that first name, so:

Q(total) = q = -Q*(2sqrt(2) + 1)/4

Or simply:

q = -Q(2*2^0.5 + 1)/4

I believe this is the same answer the book you have has. The a's go away because we could multiply each side by a^2 and each term had it in their denominator. It should kind of make sense. The amount of force needed to pull back the push each positive charge has on each other should only be dependent on the quantity of charge, yes it will be at some particular distance for that amount of charge, but ignoring the distance one can simply find the charge needed at some distance to keep the sum of any forces to be zero; so the distance doesn't need to be in there, it's more important to say it is simply some distance away, now how much charge is needed to balance out the forces upon the positive charges.

The distance is important, it is simply unneeded for this problem. They don't care how far away you place the charge, they just want to know that wherever you place it, it's value should be such that the net force acting upon any of the positive charges should be zero. Does that make sense?

Hoped this helped.

Many Smiles,
Craig

Last edited by clombard1973; Feb 19th 2010 at 05:31 AM.
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Old Feb 21st 2010, 12:22 PM   #3
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Wow...Thank you so much Craig. Ive spent hours on this problem...physics is so difficult. Your solution was really helpful i would have never figured it out. Thank you agian.
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Old Feb 21st 2010, 12:39 PM   #4
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Smile Glad I could help

Originally Posted by blob12 View Post
Wow...Thank you so much Craig. Ive spent hours on this problem...physics is so difficult. Your solution was really helpful i would have never figured it out. Thank you agian.

Hi there,
I'm glad I could be of help. And I appreciate your reply recognizing my reply. So many people just take the answers given and have no sense of appreciation for that answer regardless of the time put into answering their question for them, so I appreciate the thank you.

Many Smiles,
Craig
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