Question

An electron moves through a small distance in a uniform electric field. The magnitude of electric field is $ 2\times 10^{4}NC^{-1}. $ Now, if the direction of field is reversed keeping the magnitude same and a proton moves through the same distance, then which of the following options is correct?

• A. The time of fall will be more in case of electron
• B. The time of fall will be more in case of proton
• C. The time of fall will be same in both cases
• D. The time of fall will be independent of charge

Answer 1

Answer: (B)

The magnitude of electric field is given by
E = $ \frac{ F}{ q_0} $
or $ F = q_0 \, E $
or ma = $ q_0 E $ ( $ \because F = ma) $
a = $ \frac{ q_0 \, E }{ m} $ ...(i)
Now, distance covered by charge in electric field is given by
s = $ \frac{ 1}{ 2} at^2 $
or $ s = \frac{ 1}{ 2} \bigg( \frac{ q_0 \, E }{ m} \bigg) t^2 $ ...(ii)
[Using Eq. (i)]
Hence, time required to cover the distance s comes out to be
t = $ \sqrt{ \frac{ 2sm}{ q_0 E }} $
$ \Rightarrow t^2 \propto m $
Since, $ m_p > m_e $
Hence, a proton takes more time to cover the same distance in the uniform electric field of same magnitude.