Coulomb's Law and 4 point charges

[John Ker]

A charge Q is placed at the centre of the square of side 2.90 cm, at the corners of which four identical charges q = 6.5 C are placed. Find the value of the charge Q so that the whole system is in equilibrium. Can someone help me figure out where I have went wrong, I began by finding the force on a single corner charge with the following calculations.
2 ( k *(6.5 * 6.5)/(.029^2)) + k(6.5*6.5)/(.04101)^2

(the .04101 is my diagonal distance calculated)

This gives me 1.31E15
Where I then go 1.31E15 = kqQ/r^2

Utilizing r = .04101/2
q = q
Q = 6.5
k = the constant

I get the incorrect answer of -8.124C

Any ideas?

 

[BvU]

the force on a single corner charge

Did you add them as scalars or as vectors ?

 

[John123]

Did you add them as scalars or as vectors ?

Well I added them as scalars as each force is acting on a corner charge as a straight line? Do i need to break them up in vector addition? I am a little confused on how to start with that.

 

[Chestermiller]

Well I added them as scalars as each force is acting on a corner charge as a straight line? Do i need to break them up in vector addition? I am a little confused on how to start with that.

Is the direction of the force on a corner charge from one of the charges at an adjacent corner the same as that from a charge along the diagonal?

 

[kuruman]

Just a quick remark that this collection of charges has to be confined in the two-dimensional plane of the square. Earnshaw's theorem guarantees that there can be no equilibrium in 3d.

 

[ZapperZ]

Well I added them as scalars as each force is acting on a corner charge as a straight line? Do i need to break them up in vector addition? I am a little confused on how to start with that.

They are not acting in a straight line. The forces acting on one of the charges at a corner act in 3 different directions! Draw a free-body diagram.

For future references, even though it may be tedious to do on here, you should always accompany a question like this with a sketch.

Zz.

 

[jbriggs444]

Just a quick remark that this collection of charges has to be confined in the two-dimensional plane of the square. Earnshaw's theorem guarantees that there can be no equilibrium in 3d.

I think Earnshaw's theorem guarantees that there can be no stable equilibrium. If all of the charges are in the two dimensional plane, all forces stay within the plane and no confinement is immediately needed.

However, such an equilibrium would not be stable against small perturbations. [Like a pencil standing on its point]

 

[kuruman]

I think Earnshaw's theorem guarantees that there can be no stable equilibrium. If all of the charges are in the two dimensional plane, all forces stay within the plane and no confinement is immediately needed.

However, such an equilibrium would not be stable against small perturbations. [Like a pencil standing on its point]

Yes of course. The language of the problem "... so that the whole system is in equilibrium" suggests (at least to me) stable equilibrium. It would be better to forget equilibrium altogether and say "... so that the net electrical force on anyone charge is zero." I do not wish to detract this thread any more from the main line of thought, so I will stop here.

 

[John Ker]

Is the direction of the force on a corner charge from one of the charges at an adjacent corner the same as that from a charge along the diagonal?

It is not, but didn't I correctly account for that with my calculations?

2 ( k *(6.5 * 6.5)/(.029^2)) + k(6.5*6.5)/(.04101)^2

The equation after the addition sign I utilize Pythagoras to account for the extended distance. Is this what you mean?

 

[jbriggs444]

It is not, but didn't I correctly account for that with my calculations?

2 ( k *(6.5 * 6.5)/(.029^2)) + k(6.5*6.5)/(.04101)^2

The equation after the addition sign I utilize Pythagoras to account for the extended distance. Is this what you mean?

The point is that if you pull east with one force of, say 6.5*6.5/0.29² and north with another force of 6.5*6.5/0.29², the total force is not going to double because the two component forces are not parallel.

 

[Chestermiller]

It is not, but didn't I correctly account for that with my calculations?

2 ( k *(6.5 * 6.5)/(.029^2)) + k(6.5*6.5)/(.04101)^2

The equation after the addition sign I utilize Pythagoras to account for the extended distance. Is this what you mean?

No. The forces from the two adjacent corner charges should be multiplied by the cosine of 45 degrees. This is the component of these forces along the diagonal. (Their components normal to the diagonal cancel one another.)

 

[John Ker]

The point is that if you pull east with one force of, say 6.5*6.5/0.29² and north with another force of 6.5*6.5/0.29², the total force is not going to double because the two component forces are not parallel.

ahh thank you!

That makes a lot more sense and I just got the answer.

 

[John Ker]

No. The forces from the two adjacent corner charges should be multiplied by the cosine of 45 degrees. This is the component of these forces along the diagonal. (Their components normal to the diagonal cancel one another.)

Thank you!