Question

As shown in the figure, a dust particle with mass $m=5.0\times {10}^{-9}kg$ and charge $q_0=2.0nC$ starts from the rest at point $‘a^{\prime }$ and moves in a straight line to point $b$. What is its speed $v$ at point $b$?

• A. $26m{s}^{-1}$
• B. $34m{s}^{-1}$
• C. $46m{s}^{-1}$
• D. $14m{s}^{-1}$

Answer 1

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

Let $V_a$ and $V_b$ be velocity of particle at point $a$ and $b$ respectively.
As the particle starts from rest
$V_a=0$
Let $U_a$ be the total energy of the particle at point $a$
$U_a=P.E+K.E$
$U_a=(\frac{k\times 3\times {10}^{-9}\times q_0}{0.01}-\frac{k\times 3\times {10}^{-9}\times q_0}{0.02}$) + 0
Let $U_b$ be the total energy of the particle at point $b$
$U_b=(\frac{k\times 3\times {10}^{-9}\times q_0}{0.02}-\frac{k\times 3\times {10}^{-9}\times q_0}{0.01}$) + $\frac{1}{2}m{V}_b^2$
By total energy conservation
$U_a=U_b$
$\frac{k\times 3\times {10}^{-9}\times q_0}{0.01}-\frac{k\times 3\times {10}^{-9}\times q_0}{0.02}$ + 0 = $\frac{k\times 3\times {10}^{-9}\times q_0}{0.02}-\frac{k\times 3\times {10}^{-9}\times q_0}{0.01}$) + $\frac{1}{2}m{V}_b^2$

putting all the values $(k=9\times {10}^9,q_o=2\times {10}^9$ and $m=5\times {10}^{-9})$

Upon rearranging the terms and putting the values of $q_0$ and $m$ we get
$V_b\approx 46m{s}^{-1}$