For $\theta=30^\circ$, the result in <9/b> gives $1/3=2E_0/(E+E_0)$, hence $E=5E_0$. Acceleration from rest supplies only the change in <kinetic energy>:
$$
e|\Delta V|=E-E_0=4E_0,\qquad \boxed{|\Delta V|=\frac{4m_ec^2}{e}\simeq2.044\,\mathrm{MV}.}
$$
The accelerating <electric potential difference> is a magnitude; its sign depends on which electrode is taken as the reference, since the electron has negative <electric charge>.

Solution

Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 9 / c / Solution

 Solution (source code)