Question

The plates of a capacitor are 2⋅00 cm apart. An electron-proton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At what distance from the negative plate was the pair released?

Answer 1

Let the electric field inside the capacitor be E.

Now,
Magnitude of acceleration of the electron, $a_{\mathrm{e}}=\frac{q_{\mathrm{e}}E}{m_{\mathrm{e}}}$
Magnitude of acceleration of the proton, $a_{\mathrm{p}}=\frac{q_{\mathrm{p}}E}{m_{\mathrm{p}}}$
Let t be the time taken by the electron and proton to reach the positive and negative plates, respectively.
The initial velocities of the proton and electron are zero.
Thus, the distance travelled by the proton is given by
$x=\frac{1}{2}\frac{q_{\mathrm{p}}E}{m_{\mathrm{p}}}{t}^2$ ...(1)
And, the distance travelled by the electron is given by
$2-x=\frac{1}{2}\frac{q_{\mathrm{e}}E}{m_{\mathrm{e}}}{t}^2$ ...(2)
On dividing (1) by (2), we get
$$\begin{gathered} \frac{x}{2-x}=\frac{\left({\displaystyle \frac{q_{\mathrm{p}}E}{m_{\mathrm{p}}}}\right)}{\left({\displaystyle \frac{q_{\mathrm{e}}E}{m_{\mathrm{e}}}}\right)}=\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}=\frac{9.1\times {10}^{-31}}{1.67\times {10}^{-27}}=5.449\times {10}^{-4} \\ \Rightarrow x=10.898\times {10}^{-4}-5.449\times {10}^{-4}x \\ \Rightarrow x=\frac{10.898\times {10}^{-4}}{1.0005449}=1.08\times {10}^{-8}\mathrm{cm} \end{gathered}$$