Equation for speed of charges (electrostatics)?

[devil0150]

I am trying to do an exercise but there's no equation in the book that links speed and charges. Can anyone help? This is the exercise:

Homework Statement

There is a particle with mass = 20 grams and charge = 6 x 10^(-6) C, and another particle with mass = 50 grams and charge = -4 x 10^(-6) C. The distance between the particles is 1 m. Find the speed of each particle when their distance becomes 0.5 m.

 

[haruspex]

there's no equation in the book that links speed and charges.

Not directly, but I'm sure you have equations that relate charge and distance to force, force and mass to acceleration, acceleration and distance to speed.

 

[devil0150]

Yes I tried using coulomb's law to find the force, and then each of the accelerations (a = F/m) but to find the speed from this (v^2 = 2*a*d) I need the individual distance traveled by each particle, and I only have the sum of both distances (0.5 m).

Edit: And since the force has different value for different positions of the particles, doesn't that mean that the acceleration isn't constant? How can I find the speed using a non-constant acceleration?

 

[gneill]

Edit: And since the force has different value for different positions of the particles, doesn't that mean that the acceleration isn't constant? How can I find the speed using a non-constant acceleration?

In such cases it's often profitable to consider the problem in terms of conservation laws

 

[Nickie]

The equation for the speed of charged particles in electrostatics is given by:

v = √(2qV/m)

where v is the speed of the particle, q is the charge of the particle, V is the potential difference between the particles, and m is the mass of the particle.

In this exercise, the potential difference between the particles is given by:

V = kq1q2/d

where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q1 and q2 are the charges of the two particles, and d is the distance between them.

Substituting the given values, we get:

V = (9x10^9)(6x10^-6)(-4x10^-6)/(1m) = -2.16x10^-2 V

Now, using the above equation for speed and substituting the values for each particle, we get:

Particle 1: v1 = √(2(6x10^-6)(-2.16x10^-2)/(20x10^-3)) = 3.87x10^4 m/s

Particle 2: v2 = √(2(-4x10^-6)(-2.16x10^-2)/(50x10^-3)) = 1.55x10^4 m/s

Therefore, the speed of each particle when their distance becomes 0.5 m is 3.87x10^4 m/s and 1.55x10^4 m/s for Particle 1 and Particle 2, respectively.