[Point Charges] Can't figure this out

[Gr33nMachine]

Homework Statement

In the figure below, six charged particles surround particle 7 at radial distances of either d = 1.0 cm or 2d, as drawn. The charges are q1 = +8e, q2 = +8e, q3 = +e, q4 = +8e, q5 = +8e, q6 = +4e, q7 = +4e, with e = 1.60 10-19 C. What is the magnitude of the net electrostatic force on particle 7?---2----
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1-7-3-4
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---5----
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---6----

Homework Equations

F = k(q1)(q2)/(r^2)
k = 8.99e9

The Attempt at a Solution

I split this up into x and y components, and then figured out the sum.

For the x, ƩF = |k(q7*q1)/(r^2) - k(q7*q3)/(r^2) - k(q7*q4)/(4(r^2))|
This simplifies to: |[k(q7)/(r^2)]*[q1-q3-q4/4]|
and so ƩF = |[4ke/(.01^2)]*e(8-1-8/4)| = 4ke/.0001*5e = 20k(e^2)/.0001

I used the same process for the y component. 2 and 5 cancel out since they have the same distance and charge, so the only charge I needed to calculate was for 6. F = k(q7)(q6)/(4(r^2))
This simplifies.. F = k(6e)(4e)/((4)(.0001)) = 6k(e^2)/.0001

Now the net charge is ([20k(e^2)/.0001]^2 + [6k(e^2)/.0001]^2)^.5
Simplify this to [436(k^2)(e^4)/(10^-8)]^.5 which comes out to 4.8055e-23, which webassign says is incorrect.

 

[PeterO]

Homework Statement

In the figure below, six charged particles surround particle 7 at radial distances of either d = 1.0 cm or 2d, as drawn. The charges are q1 = +8e, q2 = +8e, q3 = +e, q4 = +8e, q5 = +8e, q6 = +4e, q7 = +4e, with e = 1.60 10-19 C. What is the magnitude of the net electrostatic force on particle 7?


---2----
---|-----
1-7-3-4
---|-----
---5----
---|-----
---6----

Homework Equations

F = k(q1)(q2)/(r^2)
k = 8.99e9

The Attempt at a Solution

I split this up into x and y components, and then figured out the sum.

For the x, ƩF = |k(q7*q1)/(r^2) - k(q7*q3)/(r^2) - k(q7*q4)/(4(r^2))|
This simplifies to: |[k(q7)/(r^2)]*[q1-q3-q4/4]|
and so ƩF = |[4ke/(.01^2)]*e(8-1-8/4)| = 4ke/.0001*5e = 20k(e^2)/.0001

I used the same process for the y component. 2 and 5 cancel out since they have the same distance and charge, so the only charge I needed to calculate was for 6. F = k(q7)(q6)/(4(r^2))
This simplifies.. F = k(6e)(4e)/((4)(.0001)) = 6k(e^2)/.0001

Now the net charge is ([20k(e^2)/.0001]^2 + [6k(e^2)/.0001]^2)^.5
Simplify this to [436(k^2)(e^4)/(10^-8)]^.5 which comes out to 4.8055e-23, which webassign says is incorrect.

I would have calculated the force between q7 and q3 [since q7 is the central charge we are analysing and q3 is just a single e. [lets call that F]

You can then express every other force in terms of that F

eg q1 is the same distance, but q1 is +8e so the force is 8F
q4 is the same size as q1, but twice the distance , so 2F [inverse square law]
etc.

 

[Gr33nMachine]

I would have calculated the force between q7 and q3 [since q7 is the central charge we are analysing and q3 is just a single e. [lets call that F]

You can then express every other force in terms of that F

eg q1 is the same distance, but q1 is +8e so the force is 8F
q4 is the same size as q1, but twice the distance , so 2F [inverse square law]
etc.

So, ƩFx = F + 8F + 8F/4 and ƩFy = 4F/4
ƩF = (11F^²+F²)^1/2
ƩF = (122F²)^1/2
F = k(4e)(e)/(.01²) = 4ke²*.01²

ƩF = 1.0168e-30

This doesn't seem right since it's 7 degrees off my previous calculations.

 

[PeterO]

So, ƩFx = F + 8F + 8F/4 and ƩFy = 4F/4
ƩF = (11F^²+F²)^1/2
ƩF = (122F²)^1/2
F = k(4e)(e)/(.01²) = 4ke²*.01²

ƩF = 1.0168e-30

This doesn't seem right since it's 7 degrees off my previous calculations.

The x-direction 8F force is in the opposite direction to the other two.

 

[Gr33nMachine]

The x-direction 8F force is in the opposite direction to the other two.

Ahh... so ƩF = (5F²+F²)^1/2 = (26F²)^1/2

This still results in a suspiciously large exponent: ƩF = 4.694e-31

 

[PeterO]

Ahh... so ƩF = (5F²+F²)^1/2 = (26F²)^1/2

This still results in a suspiciously large exponent: ƩF = 4.694e-31

Is the "real" answer of the order of 10^-22 by any chance?

 

[PeterO]

So, ƩFx = F + 8F + 8F/4 and ƩFy = 4F/4
ƩF = (11F^²+F²)^1/2
ƩF = (122F²)^1/2
F = k(4e)(e)/(.01²) = 4ke²*.01²

ƩF = 1.0168e-30

This doesn't seem right since it's 7 degrees off my previous calculations.

Looks like a divide changed to a multiply ??

 

[Gr33nMachine]

Looks like a divide changed to a multiply ??

Wow. All I can say is, stay off drugs, kids.

edit: So I ended up with a value of 1.016e-22, and it's STILL not right. I've run out of chances now, but I'm still interested in knowing what I'm doing wrong here.