Two 2.0 g plastic buttons each with + 40 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on either side of a 5.0 g button charged to + 250 nC. All three are released simultaneously.

a. How many interactions are there that have potential energy?

b. What is the final speed of the left 2.0 g plastic button? vleft=? (2.9m/s is the wrong answer)

c. What is the final speed of the right 2.0 g plastic button?

d. What is the final speed of the 5.0 g plastic button?

a. There are three potential energy interaction. b. 2.16 m/s c. 2.16 m/s d. 0 m/s

Explanation:

a. There are three potential energy interaction.

Let the charges be q₁ = + 40 nC, q₂ = + 250 nC and q₃ = + 40 nC and the distances between them be q₁ and q₂ is r, the distance between q₂ and q₃ is r and the distance between q₁ and q₃ is r₁ = 2r respectively. So, the potential energies are

U₁ = kq₁q₂/r, U₂ = kq₁q₃/2r and U₃ = kq₂q₃/r

U = U₁ + U₂ + U₃ = kq₁q₂/r + kq₁q₃/2r + kq₂q₃/r (q₁ = q₃ = q and q₂ = Q)

U = kqQ/r + kq²/2r + kqQ/r = qk/r (2Q + q/2)

b. To calculate the final speed of the left 2.0 g button, the potential energy = kinetic energy change of the particle.

ΔU = - ΔK

0 - qk (2Q + q/2) / r = - (1/2mv² - 0). Since the final potential at infinity equals zero and the initial kinetic energy is zero.

So qk (2Q + q/2) / r = - 1/2mv²

v = √[2qk (2Q + q/2) / mr] where m = 2.0 g r = 2.0 cm

substituting the values for the variables,

v = √[2 * 40 * 10⁻⁹ * 9 * 10⁹ (2 * 250 * 10⁻⁹ + 40 * 10⁻⁹/2) / 2 * 10⁻³ * 2 * 10⁻²]

v = √[360 (500 * 10⁻⁹ + 20 * 10⁻⁹) / 2 * 10⁻⁵]

v = √[720 (520 * 10⁻⁹) / 4 * 10⁻⁵] = 2.16 m/s

c. The final speed of the right 2.0 g button is also 2.16 m/s since we have the same potential energy in the system

d.

Since the net force on the 5.0 g mass is zero due to the mutual repulsion of the charges on the two 2.0 g masses, its acceleration a = 0. Since it starts from rests u = 0, its velocity v = u + at.

Hence,

v = u + at = 0 + 0t = 0 m/s