# Two 1.0 g beads, each charged to +5.0 nC are 2 cm apart. A 2.0 g bead charged to -10 nC is...

## Question:

Two 1.0 g beads, each charged to +5.0 nC are 2.0 cm apart. A 2.0 g bead charged to -10 nC is exactly halfway between them. The beads are released from the rest. What are the speeds of the positive beads, when they are 4 cm apart?

## Electric Potential energy:

Electric potential energy is a potential energy that exists between a group of charges due to their positions. It is given as being inversely proportional to the separation of charges.

The electric potential energy that exists between two point charges Q and q a distance r apart is given by

$$U \ = \ k \ \frac{Q \ q}{r}$$

where k is the Coulomb constant. Therefore, in our case, the electric potential energy before releasing the positive beads is

\begin{align*} U_{i} \ &= \ \left ( (8.98755 \ \times \ 10^9 \ N \cdot m^2/C^2) \ \times \ \left ( \frac{(+ \ 5.0 \ \times \ 10^{-9} \ C)^2}{0.02 \ m} \ + \ \frac{2 \ \times \ (+ \ 5.0 \ \times \ 10^{-9} \ C) \ \times \ (- \ 10 \ \times \ 10^{-9} \ C)}{0.01 \ m} \right ) \ \right ) \\ \\ &= \ - \ 7.86410625 \ \times \ 10^{-5} \ J \end{align*}

and the electric potential energy after the positive beads are 4 cm apart is

\begin{align*} U_{f} \ &= \ \left ( (8.98755 \ \times \ 10^9 \ N \cdot m^2/C^2) \ \times \ \left ( \frac{(+ \ 5.0 \ \times \ 10^{-9} \ C)^2}{0.04 \ m} \ + \ \frac{2 \ \times \ (+ \ 5.0 \ \times \ 10^{-9} \ C) \ \times \ (- \ 10 \ \times \ 10^{-9} \ C)}{0.02 \ m} \right ) \ \right ) \\ \\ &= \ - \ 3.93205313 \ \times \ 10^{-5} \ J \end{align*}

Therefore, the change in potential energy is

\begin{align*} \Delta U \ = \ U_{f} \ - \ U_{i} \ &= \ - \ 3.93205313 \ \times \ 10^{-5} \ J \ - \ (- \ 7.86410625 \ \times \ 10^{-5} \ J)\\ \\ &= \ 3.93205313 \ \times \ 10^{-5} \ J \end{align*}

This change in potential energy equals the change in kinetic energy of the positive beads. Each positive bead gains half the magnitude of the change in potential energy calculated above as kinetic energy.

\begin{align*} K.E \ = \ \frac{1}{2} \ \times \ 0.001 \ kg \ \times \ v^2 \ &= \ \frac{\Delta U}{2}\\ \\ \therefore \ v^2 \ = \ \frac{\Delta U}{0.001 \ kg} \ &= \ \frac{1.96602656 \ \times \ 10^{-5} \ J}{0.001 \ kg}\\ \\ &= \ 1.96602656 \ \times \ 10^{-2} \ m^2/s^2\\ \\ \therefore \ v \ &= \ \sqrt{1.96602656 \ \times \ 10^{-5} \ m^2/s^2}\\ \\ &= \ 0.1402150692 \ m/s \end{align*}

Therefore, the speed of the positive beads will be {eq}0.140 \ m/s {/eq} correct to three significant figures.