Coulomb's Law, what charge will make the two charges in static eq.

[rocapp]

Homework Statement

In the figure below the charge in the middle is
Q = -3.7 nC. For what charge
q1 will charge q2
be in static equilibrium?

Homework Equations

F = (K*q1*q2)/(r^2)

The Attempt at a Solution

I'm not sure where to go after acknowledging that Fq1->q = -Fq2->q

 

[SammyS]

Homework Statement

In the figure below the charge in the middle is
Q = -3.7 nC. For what charge
q1 will charge q2
be in static equilibrium?

Homework Equations

F = (K*q1*q2)/(r^2)

The Attempt at a Solution

I'm not sure where to go after acknowledging that Fq1->q = -Fq2->q

Let's see ...

q₁ is twice the distance from q₂ that Q is from q₂, and we have an inverse square law.

 

[rocapp]

So that means the force is (3.7^2) = ~14 nC?

 

[SammyS]

So that means the force is (3.7^2) = ~14 nC?

The unit for force is not Coulombs .

Why would you square the charge anyway ?

 

[rocapp]

I misunderstood the inverse square law.

F=q1=q2=k*q1*q2/r^2 = k*q1*Q/(r/2)^2

(9x10^9)*(q1*q2)/400 = (9x10^9)*q1*(3.7)/100

q1*q2/400 = q1*0.037
q2 = 14.8 N = F

Is that correct?

 

[SammyS]

I misunderstood the inverse square law.

F=q1=q2=k*q1*q2/r^2 = k*q1*Q/(r/2)^2

(9x10^9)*(q1*q2)/400 = (9x10^9)*q1*(3.7)/100

q1*q2/400 = q1*0.037
q2 = 14.8 N = F

Is that correct?

Doubling distance reduces force to 1/4 , so to compensate for that, charge must be 4 times what is would be at Q . (Of course with opposite sign.)

So, is 14.8 = (4)(3.7) ?

... Yes. So you're O.K.

 

[rocapp]

Oh! Wow, that's a lot simpler than I was trying to make it. Thanks!