Quick questions on Charged particles in electric fields explained: HSC Physics Module 6

Two parallel conducting plates held at a potential difference $V$ and separated by a distance $d$ produce a nearly uniform electric field in the region between them:

If the charge moves a distance $d$ in the direction of the field (or, equivalently, through a potential difference $V$ between its start and end positions):

The particle enters along the field direction (or starts at rest). Motion is one-dimensional, constant acceleration. Use $v^2 = u^2 + 2as$ with $s = d$, or use energy: $qV = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.

The particle enters horizontally between the plates with speed $v_x$, and the field is vertical. The horizontal speed is constant, the vertical motion has constant acceleration $a = qE/m$. After time $t = L / v_x$ in plates of length $L$, the vertical deflection is $y = \frac{1}{2} a t^2$.

Define the electric field strength between two parallel plates and write the equation relating it to potential difference and separation. [2 marks]

A proton is accelerated from rest through $V = 1500 \text{ V}$. Calculate its final speed. ($m_p = 1.67 \times 10^{-27} \text{ kg}$.)

An electron enters a $200 \text{ V}$, $2.0 \text{ cm}$-separation parallel plate region horizontally at $v_0 = 3.0 \times 10^7 \text{ m/s}$. (a) Find the field strength. (b) Find the transverse acceleration.