From the negative plate of a parallel plate capacitor charged to a...

Let the lower plate be at $y = 0$ and the upper plate be at $y = d$.

The electric field between the plates is $E = \dfrac{V}{d}$, directed downwards.

The magnetic field is $\vec{B} = B \hat{k}$.

The equations of motion for the electron are given by the Lorentz force:

$m_e \dfrac{d\vec{v}}{dt} = -e(\vec{E} + \vec{v} \times \vec{B})$

Considering the x-component of the force:

$m_e \dfrac{dv_x}{dt} = -e B v_y = -e B \dfrac{dy}{dt}$

Integrating with respect to time and using the initial condition $v_x = 0$ at $y = 0$:

$m_e v_x = -e B y \Rightarrow v_x = -\dfrac{e B}{m_e} y$

By the work-energy theorem, the kinetic energy of the electron at height $y$ is equal to the work done by the electric field:

$\dfrac{1}{2} m_e (v_x^2 + v_y^2) = e E y$

Substituting the expression for $v_x$:

$\dfrac{1}{2} m_e \left( \left(-\dfrac{e B}{m_e} y\right)^2 + v_y^2 \right) = e E y$

At the maximum height y_\max, the vertical component of velocity becomes zero ($v_y = 0$):

\dfrac{1}{2} m_e \left( \dfrac{e^2 B^2}{m_e^2} y_\max^2 \right) = e E y_\max

y_\max = \dfrac{2 m_e E}{e B^2}

Substituting $E = \dfrac{V}{d}$:

y_\max = \dfrac{2 m_e V}{e B^2 d}

For the electron to fail to strike the upper plate, its maximum height must be strictly less than the plate separation $d$:

y_\max < d

$\dfrac{2 m_e V}{e B^2 d} < d$

$d^2 > \dfrac{2 m_e V}{e B^2}$

$d > \left( \dfrac{2 m_e V}{e B^2} \right)^{1/2}$

Answer: $d > \left(\dfrac{2 m_e V}{e B^2}\right)^{1/2}$