An electron and a positron enter a uniform el | vedclass

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(B) The force on a charge $q$ in an electric field $E$ is given by $F = qE$. For an electron,$q = -e$,so the force is $F_e = -eE$. The acceleration is

Vedclass · Accepted Answer

$\frac{E e t^2}{m}$

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(B) The force on a charge $q$ in an electric field $E$ is given by $F = qE$. For an electron,$q = -e$,so the force is $F_e = -eE$. The acceleration is $a_e = \frac{-eE}{m}$. For a positron,$q = +e$,so the force is $F_p = +eE$. The acceleration is $a_p = \frac{eE}{m}$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,and since the initial velocity in the direction of the field is zero $(u = 0)$: Displacement of the electron in the field direction: $y_e = \frac{1}{2} a_e t^2 = -\frac{eE t^2}{2m}$. Displacement of the positron in the field direction: $y_p = \frac{1}{2} a_p t^2 = \frac{eE t^2}{2m}$. The separation distance between them in the direction of the field is $d = |y_p - y_e| = |\frac{eE t^2}{2m} - (-\frac{eE t^2}{2m})| = \frac{eE t^2}{m}$.

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 (where $\frac{e}{m}$ is the specific charge of the electron).

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