Question

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

Answer 1

Answer: 4

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