plzz solve this numerical

Since the pair of electron-proton is released, their initial velocity is zero.

Using S = ut + at²/2, and denoting by F the electrostatic force, which is qE, and same for electron and proton.

S_e = a_et²/2,          S_p = a_pt²/2,             a_e = F/m_e,               a_p = F/m_p,

Hence, S_e/S_p = a_e/a_p = m_p/m_e = 1837m_e/m_e = 1837,

S_e = 1837 S_p

S_e + S_p = 2 cm.

1837 S_p + S_p = 2 cm.

S_p = 0.001 cm = 10 μm

S_p = distance travelled by proton from it's point of release to negative plate is 10 μm.

Regards,

Team,

TopperLearning.