Finding final velocity given 2 charges


by Josh930
Tags: charges, final, velocity
Josh930
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#1
Sep16-09, 11:44 AM
P: 16
1. The problem statement, all variables and given/known data

A small metal sphere, carrying a net charge of q_1 = -2.70 \mu C, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q_2 = -7.60 \mu C and mass 1.50 g, is projected toward q_{1}. When the two spheres are 0.800 \rm m apart, q_{2} is moving toward q_{1} with speed 22.0 \rm m/s . Assume that the two spheres can be treated as point charges.



What is the speed of q2 when the spheres are 0.410 m apart?
How close does q2 get to q1?


2. Relevant equations

F=k(Q1Q2)/r^2
U=k(Q1Q2)/r




3. The attempt at a solution

I don't really know how to do this problem but i started it in this process-

The potential energy is the work done in moving the charge through the electric field. That Potential energy is donated U. So i know that the work done in the system will equal a force x distance. F = ma. ( Work= Fd). so the U = ma x distance. Then using this i could find the acceleration and the velocity... problem is i have not time intervals. so im lost. please help
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rl.bhat
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#2
Sep16-09, 11:48 AM
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What is required in the problem?
kuruman
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#3
Sep16-09, 11:50 AM
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You don't say what the question is, but it looks like you need to use conservation of mechanical energy, time is not an issue here. KE + PE at one point is equal to KE + PE at another.

Josh930
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#4
Sep16-09, 12:27 PM
P: 16

Finding final velocity given 2 charges


so i need to find the electrical potential energy. Then this is equal to the KE= 1/2mv^2;

so K(Q1Q2)/r=1/2mv^2 and solve for v at r .410m for the first question

then how do i find how close they get to each other. Using the same equations and solving for r where ke=u? if so, how do i find v^2 for the KE?
Fightfish
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#5
Sep16-09, 12:34 PM
P: 595
The kinetic energy of the charge q2 is zero at the point of closest approach since it is instantaneously at rest there.
Josh930
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#6
Sep16-09, 01:27 PM
P: 16
well i worked out part 1 of the problem and got 24.5m/s which is wrong. When finding the potential energy at the point .410, which lies between the 2 charges, do i use .410 for the distance r in the equation;

U=(8.85e9)*(-2.6e-6 * -7.6e-6)/r = 1/2mv^2 ?


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