Question

An insulating rod with linear charge density of $40\, \mu\, C / m$ and linear mass density of $0.1\, kg / m$ is released from rest in a uniform electric field $E=100\, V / m$ directed perpendicular to the rod. If rod is released and after travelling $2 \, m$ down the field, speed of rod is formed to be $(v / 10)\, m / s$, then value of $v$ is ________.

Answer 1

Taking $V=0$ at initial position or rod, then potential after ' $d$ ' distance along the field $=V=-E d$
Charge on rod $=\lambda L$
Potential energy of rod - field system initially $Q V_{\text {initial }}=0$
Potential energy of rod-field system finally
$=-\lambda L E d $ (using, $ U=q V ) $
Now using energy conservation,
$( KE + PE )_{\text {initial }}=( KE + PE )_{\text {final }}$
We get, $0=0=\frac{1}{2} m v^{2}-\lambda L E d$
$\Rightarrow v^{2}=\frac{2 \lambda L E d}{m}=\frac{2 \lambda L E d}{\mu L}=\frac{2 \lambda E d}{\mu}$
$\Rightarrow v=\sqrt{\frac{2 \lambda E d}{\mu}}=\sqrt{\left(\frac{2 \times 40 \times 10^{-6} \times 100 \times 2}{0.1}\right)}$
$=0.400 \,m / s =\frac{4}{10} m / s$
So, $ v=4$