Question

As shown in figure, a dust particle with mass $m=5.0$ $\times 10^{-9} \,kg$ and charge $q_{0}=2.0\, nC$ starts from rest at point $a$ and moves in a straight line to point $b$. What is its speed $v$ at point $b$ ?

• A. $26 \,ms ^{-1}$
• B. $34 \,ms ^{-1}$
• C. $46\, ms ^{-1}$
• D. $14 \,ms ^{-1}$

Answer 1

Answer: (C)

Apply conservation of mechanical energy between points $a$ and $b$
$( KE + PE )_{a}=( KE + PE )_{b}$
$0+\frac{k\left(3 \times 10^{-9}\right) q_{0}}{0.01}-\frac{k\left(3 \times 10^{-9}\right) q_{0}}{0.02}$
$=\frac{1}{2} m v^{2}+\frac{k\left(3 \times 10^{-9}\right) q_{0}}{0.02}-\frac{k\left(3 \times 10^{-9}\right) q_{0}}{0.01}$
Put the values and get $v=12 \sqrt{15}=46\, ms ^{-1}$