A capacitor has its plates separated by 2.00 cm . An...

Let the pair be released at a distance $x$ from the negative plate. The proton moves towards the negative plate, covering a distance $x$, while the electron moves towards the positive plate, covering a distance $d - x$, where $d = 2.00 \text{ cm}$.

The acceleration of the proton is $a_p = \dfrac{eE}{m_p}$ and the acceleration of the electron is $a_e = \dfrac{eE}{m_e}$, where $E$ is the uniform electric field between the plates.

Since both particles start from rest and take the exact same time $t$ to reach their respective plates, we can write the equations of motion as:

$x = \dfrac{1}{2} a_p t^2 = \dfrac{1}{2} \left( \dfrac{eE}{m_p} \right) t^2$

$d - x = \dfrac{1}{2} a_e t^2 = \dfrac{1}{2} \left( \dfrac{eE}{m_e} \right) t^2$

Dividing the first equation by the second yields:

$\dfrac{x}{d - x} = \dfrac{m_e}{m_p}$

$x m_p = d m_e - x m_e$

$x = \dfrac{m_e d}{m_p + m_e}$

Since $m_p \gg m_e$, the denominator can be approximated as $m_p$, giving:

$x \approx \dfrac{m_e}{m_p} d$

Substituting the standard masses $m_e = 9.1 \times 10^{-31} \text{ kg}$, $m_p = 1.67 \times 10^{-27} \text{ kg}$, and the distance $d = 2.00 \text{ cm}$:

$x \approx \dfrac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}} \times 2.00 \text{ cm}$

$x \approx 10.89 \times 10^{-4} \text{ cm}$

$x \approx 1.08 \times 10^{-3} \text{ cm}$

Answer: $1.08 \times 10^{-3} \text{ cm}$