A 4.0-g bead carries a charge of $20 \mu \mathrm { C }$. The bead is accelerated from rest through a potential difference $V$, and afterward the bead is moving at $2.0 \mathrm {~m} / \mathrm { s }$. What is the magnitude of the potential difference $V$ ?

Question

Quizplus · Accepted Answer

The kinetic energy gained by the bead is equal to the work done on it by the electric field, which is the charge times the potential difference ($qV$). The kinetic energy ($KE$) can be calculated using the formula $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. Substituting the given values, $KE = \frac{1}{2} \times 0.004 \, \text{kg} \times (2.0 \, \text{m/s})^2 = 0.008 \, \text{J}$. Since $qV = KE$, and $q = 20 \mu C = 20 \times 10^{-6} C$, solving for $V$ gives $V = \frac{KE}{q} = \frac{0.008}{20 \times 10^{-6}} = 400 \, \text{V}$.

A 40-G Bead Carries a Charge Of Of 20 20μC20 \mu \mathrm { C }20μC

A 40-G Bead Carries a Charge Of Of 20 $20 \mu \mathrm { C }$