An electron moves through a small distance in a uniform electric field. The magnitude of electric field is 2× 104NC-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?

Question

An electron moves through a small distance in a uniform electric field. The magnitude of electric field is 2× 104NC-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

Tardigrade Admin · Accepted Answer

The magnitude of electric field is given by
E =

\frac{F}{q _{0}}

or

F = q_{0} E

or ma =

q_{0} E

(

∵ F = ma)

a =

\frac{q _{0} E}{m}

...(i)
Now, distance covered by charge in electric field is given by
s =

\frac{1}{2} a t^{2}

or

s = \frac{1}{2} (\frac{q _{0} E}{m}) t^{2}

...(ii)
[Using Eq. (i)]
Hence, time required to cover the distance s comes out to be
t =

\frac{2 s m}{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.

Q. An electron moves through a small distance in a uniform electric field. The magnitude of electric field is 2×104NC−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?

The time of fall will be more in case of electron

The time of fall will be more in case of proton

The time of fall will be same in both cases

The time of fall will be independent of charge