A body of mass $1 \mathrm{~g}$ and carrying a charge $10^{-8} \mathrm{C}$ passes from two points $P$ and $Q …
A body of mass $1 \mathrm{~g}$ and carrying a charge $10^{-8} \mathrm{C}$ passes from two points $P$ and $Q . P$ and $Q$ are at electric potentials. $600 \mathrm{~V}$ and $0 \mathrm{~V}$, respectively. The velocity of the body at $Q$ is $20 \mathrm{cms}^{-1}$. It velocity in $\mathrm{ms}^{-1}$ at $P$ is
$\sqrt{0.028}$
$\sqrt{0.056}$
$\sqrt{0.56}$
$\sqrt{5.6}$
Solution
$
\begin{aligned}
m & =1 \mathrm{~g}=10^{-3} \mathrm{~kg}, \\
v & =20 \mathrm{~cm} / \mathrm{s}=20 \times 10^{-20} \mathrm{~m} / \mathrm{s} \\
q & =10^{-8} \mathrm{C} \\
V_p & =600 \mathrm{volt} \\
V_Q & =0 \text { volt }
\end{aligned}
$
Work done in moving the charge from $P$ to $Q$
$
\begin{aligned}
W & =q V_{P Q}=10^{-8}(600-0) \\
& =600 \times 10^{-8} \mathrm{~J}
\end{aligned}
$
From work energy theorem
Work done $=$ change in $\mathrm{KE}$
$
\begin{aligned}
600 \times 10^{-8} & =\frac{1}{2} m\left(20 \times 10^{-2}\right)^2-\frac{1}{2} m v^2 \\
600 \times 10^{-8} & =\frac{1}{2} \times 10^{-3} \times 4 \times 10^{-2}-\frac{1}{2} \times 10^{-3} \times v^2 \\
\frac{1}{2} \times 10^{-3} \times v^2 & =2 \times 10^{-5}-600 \times 10^{-8} \\
& =2 \times 10^{-5}-0.6 \times 10^{-5} \\
v^2 & =1.4 \times 10^{-2} \times 2=28 \times 10^{-2} \\
v & =\sqrt{28 \times 10^{-2}}=\sqrt{0.028} \mathrm{~m} / \mathrm{s}
\end{aligned}
$