Finding final velocity given 2 charges

In summary, at the point of closest approach between the charges, the kinetic energy is zero and the potential energy is 1/2mv^2.
  • #1
Josh930
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0

Homework Statement



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.

YF-23-05.jpg


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

Homework Equations



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

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 I am lost. please help
 
Last edited:
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  • #2
What is required in the problem?
 
  • #3
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.
 
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  • #4
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?
 
  • #5
The kinetic energy of the charge q2 is zero at the point of closest approach since it is instantaneously at rest there.
 
  • #6
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 ?
 

1. How do the charges affect the final velocity?

The final velocity of the charges depends on the magnitude and direction of each charge. If the charges have the same sign, they will repel each other and the final velocity of both charges will be in opposite directions. If the charges have opposite signs, they will attract each other and the final velocity of both charges will be in the same direction.

2. What is the formula for finding final velocity given 2 charges?

The formula for finding final velocity given 2 charges is v = (k*q1*q2)/m, where v is the final velocity, k is the Coulomb's constant, q1 and q2 are the charges, and m is the mass of the charged particles.

3. How is the mass of the particles taken into account in the final velocity calculation?

The mass of the particles is taken into account in the final velocity calculation through the division by the mass in the formula. This means that the final velocity will be affected by the mass of the particles, with lighter particles having a greater final velocity than heavier particles.

4. What units should be used for the charges and mass in the formula?

The charges should be in Coulombs (C) and the mass should be in kilograms (kg) in order to use the formula for finding final velocity given 2 charges. It is important to use consistent units to obtain accurate results.

5. Can the final velocity be negative?

Yes, the final velocity can be negative. This means that the charged particles are moving in the opposite direction to the initial direction of one of the charges. A negative final velocity indicates that the particles are slowing down or changing direction due to the forces exerted by the charges.

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