How Do You Calculate the Initial Separation Between Two Charged Particles?

[udontai]

Homework Statement

One particle has a mass of 3.00 10-3 kg and a charge of +8.00 µC. A second particle has a mass of 6.00 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00 10-3 kg particle is 120 m/s. Find the initial separation between the particles.

Homework Equations

let the initial separation as r and final separation as d.
kq^2/r^2 = kq^2/d^2 + 1/2m1*V^2 + 1/2m2*V^2
is the the way to solve it? how can i determine the final velocity for the 6x10^-3 particle?

The Attempt at a Solution

 

[Tanya Sharma]

Are their any external forces acting on the system comprising the two charges ?

 

[udontai]

Nope.

 

[Tanya Sharma]

What is conserved if there are no external forces acting on the system ?

 

[udontai]

momentum?

 

[Tanya Sharma]

Yes . Use Conservation of Momentum.

 

[vela]

let the initial separation as r and final separation as d.
kq^2/r^2 = kq^2/d^2 + 1/2m1*V^2 + 1/2m2*V^2
is the the way to solve it? how can i determine the final velocity for the 6x10^-3 particle?

The way you wrote the equation, you can solve it, but you'll get the wrong answer. Is the velocity for particle 1 the same as the velocity for particle 2? No. So why are you using the same variable, V, to represent those two quantities?

If you write the equation properly, you have one equations and two unknowns. You need a second equation if you hope to find specific values of each unknown.

 

[udontai]

(3.00 10-3 )x 0 + (6.00 10-3 ) x 0 = (3.00 10-3) x 120 - (6.00 10-3 ) x V
i this correct?

 

[udontai]

Yes . Use Conservation of Momentum.

(3.00 10-3 )x 0 + (6.00 10-3 ) x 0 = (3.00 10-3) x 120 - (6.00 10-3 ) x V
i this correct?

 

[Tanya Sharma]

(3.00 10-3 )x 0 + (6.00 10-3 ) x 0 = (3.00 10-3) x 120 - (6.00 10-3 ) x V
i this correct?

Correct .

kq^2/r^2 = kq^2/d^2 + 1/2m1*V^2 + 1/2m2*V^2

There is another problem with this equation apart from what Vela has pointed .

What do these terms represent ?

 

[udontai]

Correct .



There is another problem with this equation apart from what Vela has pointed .

What do these terms represent ?

as i stated r = initial distance and d= final distance.

 

[Tanya Sharma]

as i stated r = initial distance and d= final distance.

Did I ask what r and d represent ?

 

[udontai]

Did I ask what r and d represent ?

i saw the r and d is highlighted from my computer screen so i thought you asked about it. lol my bad sorry

 

[Tanya Sharma]

But you still haven't answered my question in post#10 .

 

[udontai]

But you still haven't answered my question in post#10 .

so instead of kq^2/r^2 = kq^2/d^2 + 1/2m1*V^2 + 1/2m2*V^2.
it should be kq^2/r^2 = kq^2/d^2 + 1/2m1*(V1)^2 + 1/2m2*(V2)^2 as the velocity is different.
right?

 

[Tanya Sharma]

No . Sorry but you are not answering the question . I have explicitly remarked that there is another problem with the equation apart from what Vela has pointed .

Please explain what do terms I marked in red in post#10 represent ?

 

[udontai]

kq^2/r2 represent the electrostatic force between the two charges

 

[Tanya Sharma]

kq^2/r2 represent the electrostatic force between the two charges

This is wrong .

How can you have force in conservation of energy equation ?

Instead of electrostatic force you need to have electrostatic potential energy of the two charges .

 

[udontai]

This is wrong .

How can you have force in conservation of energy equation ?

Instead of electrostatic force you need to have electrostatic potential energy of the two charges .

V = kq/r ?

 

[Tanya Sharma]

V = kq/r ?

No .

Please look up in your notes or book or on the web .

 

[udontai]

E.P.E =qV