Question

The plates of the capacitors are 2.00 cm apart. An electron-proton pair is released someone in the gap between the plates and it is found that the protons reaches the negative plate at the same time as the electron reaches the positive plates. At what distance from the negative plates was the pair released?

Answer 1

The acceleration of electron

$a_e=\frac{q_{e}E}{m_e}$

The acceleration of proton

$a_p=\frac{q_{p}E}{m_p}$

The distance travelled by proton

$x=\frac{1}{2}{a}_p{t}^2$ ___(1)

The distance travelled by electron

$2-x=\frac{1}{2}{a}_e{t}^2$ __(2)

From (1) and (2)

$\Rightarrow 2-x=\frac{1}{2}{a}_c{t}^2$ and $x=\frac{1}{2}{a}_c{t}^2$

Divide the equation (1) and (2).

$\Rightarrow \frac{x}{2-x}=\frac{a_p}{a_e}=\frac{\left(\frac{q_{p}E}{m_p}\right)}{\left(\frac{q_{e}E}{m_e}\right)}$

$=\frac{x}{2-x}=\frac{m_e}{m_p}$

$=\frac{9.1\times {10}^{-31}}{1.67\times {10}^{-27}}=\frac{9.1}{1.67}\times {10}^{-4}=5.449\times {10}^{-4}$

Therefore, simplify for value of x:

$\Rightarrow x=10.898\times {10}^{-4}-5.449\times {10}^{-4}x$

$\Rightarrow x=\frac{10.898\times {10}^{-4}}{1.0005449}=0.001089226$