Question

An electron and a positron enter a uniform electric field E perpendicular to it with equal speeds at the same time. The distance of separation between them in the direction of the field after a time 't' is
($\frac{e}{m}$ is specific charge of electron)

• A. $\frac{2Eet^2}{m}$
• B. $\frac{Eet^2}{m}$
• C. $\frac{Eet^2}{2m}$
• D. Zero

Hint:

When two particles with equal mass and opposite charge start from the same point in a uniform electric field, they accelerate in opposite directions with the same magnitude. The relative acceleration is $2a$, so separation is $\frac{1}{2}(2a)t^2 = at^2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An electron (charge $-e$, mass $m$) and a positron (charge $+e$, mass $m$) experience forces in a uniform electric field $E$. The force is given by $F = qE$. Since the field is uniform, both particles undergo constant acceleration in the direction of the force. The initial velocity in the direction of the field is zero (since they enter perpendicular to it).
Step 2: Calculate Displacement for Each Particle:
Acceleration of electron: $\vec{a}_e = \frac{-e\vec{E}}{m}$. It moves opposite to the field. Acceleration of positron: $\vec{a}_p = \frac{e\vec{E}}{m}$. It moves along the field. Since they have the same mass magnitude and charge magnitude, the magnitude of acceleration is $a = \frac{eE}{m}$. Using the kinematic equation $s = ut + \frac{1}{2}at^2$ with $u=0$ in the direction of the field: Displacement of positron in time $t$: $y_p = \frac{1}{2} a t^2 = \frac{1}{2} \left(\frac{eE}{m}\right) t^2$ (along field). Displacement of electron in time $t$: $y_e = \frac{1}{2} a t^2 = \frac{1}{2} \left(\frac{eE}{m}\right) t^2$ (opposite to field).
Step 3: Calculate Separation:
Since they move in opposite directions along the field line, the total separation distance is the sum of their individual displacements. Separation $d = y_p + y_e = \frac{1}{2} \left(\frac{eE}{m}\right) t^2 + \frac{1}{2} \left(\frac{eE}{m}\right) t^2 = \left(\frac{eE}{m}\right) t^2$. Final Answer:
$\frac{Eet^2}{m}$.