The electric potential difference $Delta V$ is related to the change of the electric potential energy $Delta PE$ by the relationship $Delta PE = qDelta V$. Solving for $q$ and substituting the known values, we find

$$
q = frac{Delta PE}{Delta V} = frac{1.37times10^{-15};mathrm{J}}{2140;mathrm{V}} = 6.40times10^{-19};mathrm{C}
$$

where we used 1 J = (1 C)(1 V).

$6.40times10^{-19}$ C

As the electrons are accelerated between the two plates, the electric potential energy is converted to a kinetic energy gained by the electrons. Thus we have $Delta KE = Delta PE$. Since we know $Delta PE = eDelta V$, where $Delta V$ is the voltage between the plates and $e$ is the magnitude of the electron’s charge, we obtain $Delta KE = eDelta V$. Solving for $Delta V$ and substituting the known values (remember that 1 J = (1 C)(1 V)), we find

$$
Delta V = frac{Delta KE}{e} = frac{4.1times10^{-15};mathrm{J}}{1.6times10^{-19};mathrm{C}} = 26;mathrm{kV}
$$

$$
26;mathrm{kV}
$$