A particle (q = 3.0 mC, m = 20 g) has a speed of 20 m/s when it enters a region where the electric field has a constant magnitude of 80 N/C and a direction which is the same as the velocity of the particle. What is the speed of the particle 3.0 s after it enters this region?
A) 68 m/s
B) 44 m/s
C) 56 m/s
D) 80 m/s
E) 36 m/s

Question

Quizplus · Accepted Answer

The force on the particle due to the electric field is given by $F = qE$, where $q$ is the charge of the particle and $E$ is the electric field strength. Substituting the given values, $F = 3.0 \times 10^{-3} \, \text{C} \times 80 \, \text{N/C} = 0.24 \, \text{N}$. The acceleration $a$ of the particle can be found using Newton's second law, $F = ma$, where $m$ is the mass of the particle. Given that $m = 20 \, \text{g} = 0.02 \, \text{kg}$, we find $a = \frac{F}{m} = \frac{0.24 \, \text{N}}{0.02 \, \text{kg}} = 12 \, \text{m/s}^2$. The final speed $v$ of the particle after time $t = 3.0 \, \text{s}$ can be found using the equation $v = u + at$, where $u = 20 \, \text{m/s}$ is the initial speed. Substituting the values, $v = 20 \, \text{m/s} + (12 \, \text{m/s}^2)(3.0 \, \text{s}) = 56 \, \text{m/s}$.

A Particle (Q = 3