Answer
$-2.0\times10^5V/m$ to $2.0\times10^5V/m$
Work Step by Step
When the electric field is 0, the beam will not be deflected and will reach the center of the screen.
When going through the deflection plates, electrons will be accelerated in either the upward or downward direction for distance $y_1$. After they pass the deflection plates, they will travel at constant velocity for distance $y_2$.
$y_1=\frac{1}{2}a_yt_1^2$
$y_2=v_yt_2=v_y\frac{0.32m}{v_x}$
$y_1+y_2=0.15m$
Kinetic energy gained is potential energy lost, so $\frac{1}{2}mv_x^2=qV$
$v_x=\sqrt{\frac{2qV}{m}}$
From $F=Eq$; $ma_y=Eq$
$a_y=\frac{Eq}{m}$
$t_1=\frac{0.026m}{v_x}$
$v_y=v_{yi}+a_yt_1=0+E(0.026m)\sqrt{\frac{q}{2mV}}$
$\frac{1}{2}\frac{Eq}{m}(\frac{0.026m}{\sqrt{\frac{2qV}{m}}})^2+(E(0.026m)\sqrt{\frac{q}{2mV}})\frac{0.32m}{\sqrt{\frac{2qV}{m}}}=0.15m$
$E=\frac{2(6\times10^3V)(0.30m)}{(0.026m)^2+2(0.026m)(0.34m)}=2.0\times10^5V/m$
For the electron to travel in the opposite direction for the same distance, it must be deflected by an electric field of the same strength but opposite direction.
