A beam of electrons of velocity $3×10^7\,ms^{-1}$ is deflected 1.5 mm in passing 10 cm through an electric field of 1800 V $m^{-1}$ perpendicular to their path. The value of e/m for electron is:

$1.78×10^{11}C\,kg^{-1}$ $2×10^{11}C\,kg^{-1}$ $1.5×10^{11}C\,kg^{-1}$ $3.5×10^{11}C\,kg^{-1}$

$1.5×10^{11}C\,kg^{-1}$

$y=\frac{1}{2}at^2=\frac{1}{2}\frac{eE}{m}\frac{x^2}{v^2}$ or $\frac{e}{m}=\frac{2yv^2}{Ex^2}=\frac{2×1.5×10^{-3}×(3×10^7)^2}{1800×(0.1)^2}=1.5×10^{11}C\,kg^{-1}$