Question

As shown in Fig, a dust particle with mass $m=5.0\times {10}^{-9}kg$ and charge $q_0$= 2.0 n C 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 m$s^{-1}$
• B. 34 m$s^{-1}$
• C. 46 m $s^{-1}$
• D. 14 m $s^{-1}$

Answer 1

The correct option is C 46 m $s^{-1}$

There is no external force so apply conservation of mechanical energy between points a and b
initially particle is at rest so kinetic energy at point a is zero
$(KE+PE{)}_a=(KE+PE{)}_b$
$0+\frac{k(3\times {10}^{-9})q_0}{0.01}-\frac{k(3\times {10}^{-9})q_0}{0.02}=\frac{1}{2}m{v}^2+\frac{k(3\times {10}^{-9})q_0}{0.02}-\frac{k(3\times {10}^{-9})q_o}{0.01}$
Put the values and get $v=12\sqrt{15}=46m{s}^{-1}$