A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 55000m/s . What will be the final speed of an electron released from rest at the negative plate?

Question

anonymous · Answer

Let the potential difference between the plates be V
By energy conservation: 
qV=1/2*m_p*v_p^2                   where m_p= mass of proton, v_p= velocity of proton on reaching the opposite plate
or v_p=sqrt( 2qV/m_p)=55000

By the same principle,
qV=1/2*m_e*v_e^2   where, m_e=mass of electron  v_e=velocity of electron on reaching the opposite plate

or v_e=sqrt(2qV/m_e)

Now, v_e/v_p=sqrt(m_p/m_e)
 Or,    v_e=55000*sqrt(m_p/m_e)
          v_e=55000*sqrt(1837) .... since m_p/m_e=1837 approx.