How Fast Are Charged Particles Moving When Halfway Together?

[metalmagik]

Homework Statement

Two particles each have a mass of 5.9e-3 kg. One has a charge of +5.0e-6 C, and the other has a charge of -5.0e-6 C. They are initially held at rest at a distance of 0.70 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-half its initial value?

Homework Equations

$$F = kq1q2/r^2$$

$$E = F/q$$

$$V=Ed$$

$$CV = q$$

$$Energy = 1/2CV^2$$

$$KE = 1/2mv^2$$

The Attempt at a Solution

F = kq1q2/r^2

F = (8.99e9)(5e-6)(5e-6)/.35^2

F = 1.83 N

E = F/q

E = 1.83/5e-6

E = 366,000 N/C

V = Ed

V = (366000)(.35)

V = 128,100 V

q = CV

(5e-6) = (128,100)C

C = 3.90e-11

Energy = 1/2CV^2

Energy = .5(3.9e-11)(128100)^2

Energy = .32 J

KE = 1/2mv^2

.32 = 1/2(5.9e-3)v^2

v = 10.4 m/s

Verification on this would be very greatly appreciated.

 

[OlderDan]

Several of the equations you have written are not relevant to this problem. One key equation you have left out and need to use is the equation for the electrical potential energy of two point charges exerting forces on one another. That energy can be thought of as the potential energy of one charge that finds itself in the electric field produced by another charge. Associated with the vector electric field is a scalar electric potential, which is potential energy per unit charge. What is the electric potential produced by a point charge?

 

[m2003]

Your solution looks correct. To verify, we can use the equation for potential energy: PE = kq1q2/r. At the initial separation of 0.70m, the potential energy between the two particles is 1.83 J. When the separation is reduced to 0.35m, the potential energy becomes 3.66 J. This is equal to the sum of the kinetic energies of the two particles, which is 2(KE) = 2(0.32 J) = 0.64 J. This confirms that the particles are moving with a speed of 10.4 m/s when the separation is one-half its initial value. Great job on your solution!