Question #9baf6

May 10, 2016

$5.7 \times {10}^{5} m {s}^{-} 1$, rounded to first decimal place.

Explanation:

When a charged particle having charge $q$ is situated in an electric field $\vec{E}$, the force $\vec{F} = q \cdot \vec{E}$

Considering only the modulus part $| \vec{F} | = q \cdot | \vec{E} |$
We also know from Newton's second law that Force $\vec{F} = m \cdot \vec{a}$, where $\vec{a}$ is the acceleration.
again considering the modulus part
$| \vec{F} | = m \cdot | \vec{a} |$
Equating both we get
$m \cdot | \vec{a} | = q \cdot | \vec{E} |$
or $| \vec{a} | = \frac{q}{m} \cdot | \vec{E} |$
Kinematic equation for a particle moving with constant acceleration is
${v}^{2} - {u}^{2} = 2 a s$,
where $v$ is the final velocity, $u$ is the initial velocity and $s$ is the distance traveled. Inserting given values and solving for $v$

${v}^{2} - {0}^{2} = 2 \frac{q}{m} \cdot | \vec{E} | s$
or ${v}^{2} = 2 \times \frac{3.2 \times {10}^{-} 19}{6.4 \times {10}^{-} 27} \cdot \left(1.6 \times {10}^{5}\right) \times \left(2 \times {10}^{-} 2\right)$
or $v = \sqrt{3.2 \times {10}^{11}}$
or $v = 5.7 \times {10}^{5} m {s}^{-} 1$, rounded to first decimal place.