Electric Potential Energy Video Lessons

Concept

Problem: Two 2.0-mm diameter beads, C and D, are 12 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 2.5 nC. Bead D has mass 1.7 g and charge -1.0 nC. If the beads are released from rest, what is the speed vC at the instant the beads collide? What is the speed vD at the instant the beads collide?

FREE Expert Solution

Electrostatic potential energy:

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}}$

Momentum:

$\overline{){\mathbit{p}}{\mathbf{=}}{\mathbit{m}}{\mathbit{v}}}$

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}}$

Law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$ Wnc is the work done by non-conservative forces like friction.

We'll use conservation of momentum to determine the relationship in the velocities of the balls. Take bead C is moving in the positive direction and bead D is moving in the negative direction.

pi = pf

0 = mcvc - mdvd

mcvc = mdvd

(1.0g)vc = (1.7g)vd

vc = 1.7vd

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Problem Details

Two 2.0-mm diameter beads, C and D, are 12 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 2.5 nC. Bead D has mass 1.7 g and charge -1.0 nC. If the beads are released from rest, what is the speed vC at the instant the beads collide? What is the speed vD at the instant the beads collide?