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What amount of upward electron field is required to balance an electron. If an electron falls from a height of \[0.015\] m in an electric field. If the field is reversed then the proton will fall from the same height. Find the time of fall in both cases also compare the observation with gravitation given \[E=2\times {{10}^{4}}\text{ }N/m.\]

Answer
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Hint: We know that the electron falls from rest, so its initial velocity is zero.The electric field is present in the region, so the electron will experience electric force.The vertical distance to be covered is $h$. The electric field magnitude is $E$ and it is directed upwards. Since the electron is negatively charged, it will experience force downwards.

Complete answer:
As we know that any charged particle in a region of electric field experiences electric force which does not cause the charge to be in motion. Both electron and proton have the same magnitude of charge but of opposite polarity. The charge on protons is positive and the charge on electrons is negative. Both the proton and electron experience electric force. Also, in this problem field was from down to top, the force experienced by the electron was downwards and that by the proton was upwards. Here we have $mg=eE$ which implies that $E=mg/e.$

On substitution we get; $\dfrac{9.1\times {{10}^{-31}}\times 9.8}{1.6\times {{10}^{-19}}}=5.57\times {{10}^{-11}}N/C$ and for $mg=eE$ for balancing electric field with gravity. Similarly for electron if the field is reversed and carried out by $ma=mg+eE$ since \[eE=mg\] the above equation can be rewritten as $mg+mg.$

Now, acceleration can be given by; $a=\dfrac{2mg}{m}$ after cancelling both masses from numerator and denominator we get final acceleration of electrons as $2g.$Now to identify the time for fall for proton $d=\sqrt{\dfrac{2h}{2g}}=\sqrt{\dfrac{h}{g}}$ on further substitution we get $\sqrt{\dfrac{2\times 0.015}{2\times 9.8}}=0.391s$

Therefore, the time of fall in both cases is $0.391\,s.$

Note: Remember that any charge when placed in an external electric field will experience an electric force. In this case the force experienced by all the electrons in the object should balance the weight.