= Solution
The magnitude of the electric force is $f=e\mathcal E$, directed so that the electron gains <energy>. The work across the gap is $e\mathcal E L=50\,\mathrm{MeV}$. With the source precision $mc^2=0.511\,\mathrm{MeV}$,
$$
\boxed{\gamma=1+\frac{50}{0.511}=98.84736,\qquad
\beta=\sqrt{1-\gamma^{-2}}=0.999948826.}
$$
Reducing the field by 20 percent gives kinetic <energy> $40\,\mathrm{MeV}$ and
$$
\gamma_{80\%}=79.27789,\qquad\beta_{80\%}=0.999920442,
\qquad\boxed{\Delta\beta=-2.83837\times10^{-5}.}
$$
A large fractional energy change therefore produces only a small speed change, since the original electron is already highly relativistic.
For the transit time, combine the distance formula from <1/d/solution> with $\gamma^2-1=[ft/(mc)]^2$:
$$
\boxed{t=\frac{mc}{e\mathcal E}\sqrt{\gamma^2-1}
=\frac Lc\sqrt{\frac{\gamma+1}{\gamma-1}}
=3.36956\times10^{-8}\,\mathrm s.}
$$
This is slightly longer than $L/c$, since the electron starts from rest. For the reduced field it becomes $3.37798\times10^{-8}\,\mathrm s$.
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