Solution:
Two forces will act on the tank .
(a) Electrostatic force (b) Thrust force
Let v be the velocity at any instant . Then
$$F_{net}=QE-mvn$$
$$(m_{0}+mnt)\;\large\frac{dv}{dt}=QE-mnv$$
$$\int_{0}^{v}\;\large\frac{dV}{QE-mnv}=\int_{0}^{t}\;\large\frac{dt}{m_{0}+mnt}$$
$$ln(\large\frac{QE}{QE-mnv})=ln(\large\frac{m_{0}+mnt}{m_{0}})$$
$$m_{0}QE=m_{0}QE-m_{0}mnv+QEmnt-m^2n^2vt$$
$$v=\large\frac{QEt}{m_{0}+mnt}\;.$$
