To solve this problem, we will use the concepts of motion under uniform acceleration and the properties of electric forces impacting different particles.

First, let's recall that the force exerted by an electric field $E$ on a charge $q$ is given by $F = qE$. Both the electron and the proton experience a force due to the electric field E, but in opposite directions because their charges have opposite signs. However, since the magnitude of their charges is equal ($|e|$ for both), the magnitude of the force they experience is the same and is given by $F = eE$, where $e$ is the elementary charge.

Next, let's apply Newton's second law, $F = ma$, where $m$ is the mass of the particle and $a$ is its acceleration. From this, we can solve for $a$ and find that the acceleration $a$ experienced by each particle is directly proportional to the force exerted and inversely proportional to the mass of the particle, $a = \frac{F}{m} = \frac{eE}{m}$.

Since both particles start from rest and move under uniform acceleration, we can use the equation of motion:

$s = ut + \frac{1}{2}at^2$

Given that the initial velocity $u = 0$ (they start from rest), this simplifies to:

$s = \frac{1}{2}at^2$

Substituting $a = \frac{eE}{m}$ gives:

$s = \frac{1}{2}\left(\frac{eE}{m}\right)t^2$

Thus, solving for $t^2$, we get:

$t^2 = \frac{2ms}{eE}$

which implies that:

$t = \sqrt{\frac{2ms}{eE}}$

We can now write the times for the electron ($t_e$) and the proton ($t_p$) using their respective masses ($m_e$ for electron and $m_p$ for proton):

$t_e = \sqrt{\frac{2m_es}{eE}}$

$t_p = \sqrt{\frac{2m_ps}{eE}}$

The question asks us for the ratio $t_e / t_p$, which can be calculated as follows:

$\frac{t_e}{t_p} = \frac{\sqrt{\frac{2m_es}{eE}}}{\sqrt{\frac{2m_ps}{eE}}} = \sqrt{\frac{m_e}{m_p}}$

Therefore, the correct answer to the question "If the time taken by them to cover that distance is $t_e$ and $t_p$ respectively, then $t_e / t_p$ is equal to" is:

Option B:

$\sqrt{\left(\frac{m_e}{m_p}\right)}$

Electrostatics

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