# [Point Charges] Can't figure this out

• Gr33nMachine
In summary: I've tried a number of different variations to no avail.Is the "real" answer of the order of 10-22 by any chance?In summary, the net electrostatic force on particle 7, surrounded by six charged particles at radial distances of either d = 1.0 cm or 2d, is calculated by splitting the forces into x and y components and summing them. The resulting value is then found to be incorrect and further attempts to calculate the force using different methods also result in incorrect values. The correct answer may be of the order of 10^-22.
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.

Gr33nMachine said:

## 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.

PeterO said:
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.

Gr33nMachine said:
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.

PeterO said:
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

Gr33nMachine said:
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?

Gr33nMachine said:
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 ??

PeterO said:
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.

## 1. How do point charges affect electric fields?

Point charges have an electric field that radiates outward from the charge. The strength of the electric field decreases as the distance from the charge increases.

## 2. Can point charges have a negative charge?

Yes, point charges can have a negative charge. This means that they have an excess of electrons compared to protons.

## 3. How do point charges interact with each other?

Point charges can either attract or repel each other, depending on the sign of their charges. Like charges (positive-positive or negative-negative) repel each other, while opposite charges (positive-negative) attract each other.

## 4. How do point charges affect the potential energy of a system?

Point charges contribute to the potential energy of a system by creating an electric potential. The potential energy increases as the distance between the charges decreases, and decreases as the distance increases.

## 5. Can point charges exist in isolation?

No, point charges cannot exist in isolation. They are always surrounded by an electric field, which influences other charges in the vicinity.

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