Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 1 / e / 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 $m{c}^2=0.511\mathrm{MeV}$,Reducing the field by 20 percent gives kinetic energy $40\mathrm{MeV}$ andA 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 solution with ${\gamma }^2-1=[ft/(mc){]}^2$:This is slightly longer than $L/c$, since the electron starts from rest. For the reduced field it becomes $3.37798\times 10^{-8}\mathrm{s}$.