Net Force on Center Charge in Y-Direction in Square of Charges

[Winzer]

Homework Statement

Four charges, Aq, Bq, Cq, and Dq (where q = 3.50 ×10−7 C), sit in a plane at the corners of a square whose sides have length d = 23.5 cm, as shown in the diagram below. A charge, Eq, is placed at the origin at the center of the square. (Note that A, B, C, D, and E are integer multipliers.)

DATA: A = 2, B = 4, C = 5, D = 8, E = 1. Consider the charge at the center of the square, Eq.
What is the net y-component of the force on this charge?

Consider the situation where A=B=C=D=1, E=-1, and the following statements. Select "True" or "False" for each statement.
1) The sum of the forces on the center charge in the y-direction equals zero.
2)If one were to triple the magnitude of the negative charge, the negative charge would be in equilibrium.
3)The equilibrium point at the center is a stable equilibrium for the motion of the negative charge in the plane of the square.
4)The sum of the forces on the center charge in the x-direction does not equal zero.
5)If one were to double the magnitude of the upper-right-hand positive charge, the negative charge would be in equilibrium.
6)The equilibrium point at the center is an unstable equilibrium for the motion of the negative charge in the line from the center charge and perpendicular to the plane of the square.

Homework Equations

$$F= \frac{Kq_{1}q_{2}}{r^2}$$

The Attempt at a Solution

So for the first one I break up each each force acting on E. I then do $$Fsin(\theta)=F_{y}$$ for each one and take the sum right, i keep getting it wrong for some reason.
For the second part:
1)T
2)T
3)T
4)F
5)F
6)F
Thanks

 

[rootX]

is it 0.197N?

 

[Winzer]

is it 0.197N?

How did you get that? I got 0.1725
What did you get fi the y components of each charge?

 

[rootX]

How did you get that? I got 0.1725
What did you get fi the y components of each charge?

So, first find E (net Electric field) at point E.

what was your r distance? - it should be same for all points
and write charge values like p(q) where p is the integer and q is the q value
and sin(theta) for all is same = 1/sqrt(2)

so, net E = sin(thea)*k*q*(1/r^2)[p[A]+p+p[C]+p[D]]
factored out everything that was common.

so try it.

Edit: from net E I meant E in y direction

 

[Winzer]

so y didn't you use $$F= \frac{Kq_{1}q_{2}}{r^2}$$ on each pair of charges?

 

[rootX]

so y didn't you use $$F= \frac{Kq_{1}q_{2}}{r^2}$$ on each pair of charges?

you can also do that, but I preferred to find electric field first, it really doesn't matter much

 

[Winzer]

Oh ok. What do you think about the T/F questions?

 

[rootX]

I didn't get most of them ><.

like
"2)If one were to triple the magnitude of the negative charge, the negative charge would be in equilibrium."

I don't know where's the negative charge.

And 1) shouldn't it be F, we just found a value for it?

Edit: oops.. I missed that sentence..so wait for like a sec

Edit3: I don't know #3 amd #6, but for others I also got the same answers.

 

[Winzer]

See that is where I am stuck also.

 

[Winzer]

So any ideas on the true/false?

 

[rootX]

nopes

 

[Winzer]

anyone? pretty confident on the true false except for 3 &6. Aren't they both kind of asking the same thing though? The central charge E is stable in its plane and move out of it?

 

[learningphysics]

This is all for A=B=C=D=1 and E=-1

1) The sum of the forces on the center charge in the y-direction equals zero. T

2)If one were to triple the magnitude of the negative charge, the negative charge would be in equilibrium. T

3)The equilibrium point at the center is a stable equilibrium for the motion of the negative charge in the plane of the square. T

4)The sum of the forces on the center charge in the x-direction does not equal zero. F

5)If one were to double the magnitude of the upper-right-hand positive charge, the negative charge would be in equilibrium. F

6)The equilibrium point at the center is an unstable equilibrium for the motion of the negative charge in the line from the center charge and perpendicular to the plane of the square. F

 

[learningphysics]

Hmmm... I'm getting
7*sin(45)kq^2/r^2
=7*0.7071068*9E9(3.5E-7)^2/(0.235/2)^2
=0.395N for the first part...

Oops... should be:

7*sin(45)kq^2/r^2
=7*0.7071068*9E9(3.5E-7)^2/[(0.235/2)^2]*2
=0.1976N for the first part...

 

[Winzer]

yah first part is right but T & F is wrong.

 

[learningphysics]

yah first part is right but T & F is wrong.

Oops... yeah, 3 and 6 should be opposite... 3 is false. 6 is true. Because when the negative charge is moved from the center, it will be attracted towards one end...

 

[learningphysics]

yah first part is right but T & F is wrong.

0.395N is right?

 

[Winzer]

The answer to the first part is .197N. Second part is still wrong.

 

[learningphysics]

The answer to the first part is .197N. Second part is still wrong.

Oops... both 3 and 6 are false...

The negative charge is in unstable equilibrium for motion within the plane... but it is in stable equilibrium for motion perpendicular to the plane...

Reason is it will be attracted back downwards towards the plane, back to the equilibrium point...

 

[learningphysics]

The answer to the first part is .197N. Second part is still wrong.

Yeah, I messed up... now I get 0.1976N...

 

[Winzer]

mmmm...one or more are still wrong. I think 6 is false too as well as 3.

 

[learningphysics]

mmmm...one or more are still wrong. I think 6 is false too as well as 3.

Check the true false questions to make sure you posted exactly as they're written...

 

[Winzer]

ok so it is
1.T
2.T
3.F
4.F
5.F
6.F
But can u clarify why 6 is F?

 

[learningphysics]

ok so it is
1.T
2.T
3.F
4.F
5.F
6.F
But can u clarify why 6 is F?

6 is false, because when the center charge moves up above the plane, the other four charges will exert a net downward force... and this force tries to bring the center charge back down towards the plane... that's stable equilibrium, not unstable equilibrium...

It's stable when the charge is pulled back to the equilibrium point... it's unstable when it is pushed away from the equilibrium point...

 

[Winzer]

ok, it is obvious now, thanks

 

[learningphysics]

ok, it is obvious now, thanks

No prob... but at least one of them is still wrong isn't it? I don't see which ones...

 

[Winzer]

Hmmm... I'm getting
7*sin(45)kq^2/r^2
=7*0.7071068*9E9(3.5E-7)^2/(0.235/2)^2
=0.395N for the first part...

Oops... should be:

7*sin(45)kq^2/r^2
=7*0.7071068*9E9(3.5E-7)^2/[(0.235/2)^2]*2
=0.1976N for the first part...

Wait, how did you get to that?

 

[learningphysics]

Do you see how each charge contributes:

N*sin(45)kq^2/r^2 in the vertical direction... where N is the multiplier of the outside charge... the multiplier for E = 1...

for example Aq exerts a force on 1q of Akq^2/r^2... then the vertical component of that force is Asin45kq^2/r^2

 

[Winzer]

but y the q^2, shouldn't it be $$q_{1}q_{2}$$?

edit: oh wait i see what you did. shouldn't r=.332?

 

[learningphysics]

but y the q^2, shouldn't it be $$q_{1}q_{2}$$?

yeah, but q1 = Aq... and q2 = Eq = 1q

So you get q1q2 = Aq^2

 

[learningphysics]

but y the q^2, shouldn't it be $$q_{1}q_{2}$$?

edit: oh wait i see what you did. shouldn't r=.332?

I'm getting r^2 = (0.235/2)^2 + (0.235/2)^2

 

[Winzer]

mmm...so what is wrong with:
$$F= \frac{qKCos(\theta)}{r^2}[-1.4e^-6-1.75e^-6+7e^-7+2.8e^-6]$$
i chose the negitives because -x direction.

edit: for the x direction

 

[learningphysics]

mmm...so what is wrong with:
$$F= \frac{qKCos(\theta)}{r^2}[-1.4e^-6-1.75e^-6+7e^-7+2.8e^-6]$$
i chose the negitives because -x direction.

edit: for the x direction

Can you show how you get your numbers? I'm not getting those numbers...

 

[Winzer]

well A=2*3.50e-7=7.0e-7
B=4*3.50e-7=1.4e-6...

 

[learningphysics]

well A=2*3.50e-7=7.0e-7
B=4*3.50e-7=1.4e-6...

Oh, I see now... Ok... but you should be taking the y-components, not the x-components... so the 1.4e-6 and 7e-7 should have the same sign. And the other two should have the same sign.