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 2m down the field, speed of rod is found to be (v/10) m/s, then value of v is ______

Taking potential V_initial = 0 at initial position of rod, then potential after travelling distance 'd’ is V=-Ed
Charge on rod = $\lambda$L
Potential energy of rod-field system initially QV_initial=0
Potential energy of rod-field system finally

$=-\lambda LEd\text{ (using, }U=qV)$

Now using energy conservation,

$\begin{array}{l}(\mathrm{KE}+\mathrm{PE}{)}_{\text{initial }}=(\mathrm{KE}+\mathrm{PE}{)}_{\text{final }}\\ \text{ We get, }0+0=\frac{1}{2}m{v_0}^2-\lambda LEd\\ \Rightarrow {v_0}^2=\frac{2\lambda LEd}{m}=\frac{2\lambda LEd}{\mu L}=\frac{2\lambda Ed}{\mu }\\ \begin{array}{rl}\Rightarrow v_0& =\sqrt{\frac{2\lambda Ed}{\mu }}=\sqrt{\left(\frac{2\times 40\times {10}^{-6}\times 100\times 2}{0.1}\right)}\\ & =0.400\mathrm{m}/\mathrm{s}=\frac{4}{10}\mathrm{m}/\mathrm{s}\end{array}\\ \text{ So, }v=4\end{array}$