The plate of a capacitor are 2.00cm 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 what distance from the negative plate was the pair released?

Explanation

The accn. Of electron `= 1_e `
The accln. Of the Proton `=a_p`
The dist. Travelled by prton ` x = a_p t ^(2) `
`2 - x = a_e t^(2) `
`rArr x = 10. 898 xx 10^(4) `
`- 5.449 xx 10^(-4) x`
`rArr x + 0.0005449 x = 1.0005449`
`10.898 xx 10^(-4) `
`x = 0.001089226 xx 10^(4) cm.`