Solution

Let $U$ denote the accelerating potential difference and $V$ the minimum potential difference between the capacitor plates.

After being accelerated through the potential difference $U$, the total relativistic energy of the electron is

$$E=m_ec^2+eU.$$

The relativistic energy-momentum relation is

$$E^2=p^2c^2+m_e^2c^4.$$

Hence

$$p^2c^2=E^2-m_e^2c^4.$$

The electric field inside the capacitor is directed perpendicular to the plates. Therefore, the component of the electron momentum parallel to the plates remains constant:

$$p_{\parallel}=p\cos\alpha.$$

The minimum potential difference corresponds to the limiting case in which the electron just reaches the upper plate with zero momentum component perpendicular to the plates. Therefore, its momentum at that point is

$$p'=p\cos\alpha.$$

The total energy of the electron at the upper plate is then

$$E'=\sqrt{m_e^2c^4+p^2c^2\cos^2\alpha}.$$

Using the expression for $p^2c^2$, we obtain

$$E'=\sqrt{m_e^2c^4+\left(E^2-m_e^2c^4\right)\cos^2\alpha}.$$

After simplification,

$$E'=\sqrt{E^2\cos^2\alpha+m_e^2c^4\sin^2\alpha}.$$

As the electron moves from the lower plate toward the upper plate, its energy decreases by $eV$. Therefore,

$$E'=E-eV.$$

Thus,

$$eV=E-\sqrt{E^2\cos^2\alpha+m_e^2c^4\sin^2\alpha}.$$

Substituting $E=m_ec^2+eU$, we get

$$eV=m_ec^2+eU-\sqrt{\left(m_ec^2+eU\right)^2\cos^2\alpha+m_e^2c^4\sin^2\alpha}.$$

Dividing by $e$, we obtain

$$V=U+\frac{m_ec^2}{e}-\sqrt{\left(U+\frac{m_ec^2}{e}\right)^2\cos^2\alpha+\left(\frac{m_ec^2}{e}\right)^2\sin^2\alpha}.$$

Now we substitute the numerical values:

$$U=1\,\mathrm{MV},\qquad \alpha=30^\circ,\qquad \frac{m_ec^2}{e}=0.511\,\mathrm{MV}.$$

Therefore,

$$V=1+0.511-\sqrt{(1+0.511)^2\cos^230^\circ+0.511^2\sin^230^\circ}.$$

Since

$$\cos^230^\circ=\frac34,\qquad \sin^230^\circ=\frac14,$$

we obtain

$$V=1.511-\sqrt{1.511^2\cdot\frac34+0.511^2\cdot\frac14}.$$

Hence,

$$V\approx0.178\,\mathrm{MV}.$$