Question 823: Electrostats & Capacitance

Question

An electron of mass ' $m$ ' and charge ' $q$ ' is accelerated from rest in a uniform electric field of intensity ' $E$ '. The velocity acquired by it as it travels a distance ' $l$ ' is ' $v$ '. The ratio $\frac{\mathrm{q}}{\mathrm{m}}$ in terms of $E, l$ and $v$ is

Accepted Answer

When an electron with mass 'm' and charge 'q' is accelerated from rest in a uniform electric field with intensity 'E', it attains a velocity 'v' while traveling a distance 'l'. We need to find the expression for the ratio $\frac{q}{m}$ in terms of $E$, $l$, and $v$.
First, we use the relationship between force and acceleration:
$ F = ma \quad 	ext{and} \quad F = qE $
Thus, we equate these expressions:
$ qE = ma $
So, the acceleration $a$ is given by:
$ a = \frac{qE}{m} \quad 	ext{...(i)} $
Next, we apply the equation of motion for constant acceleration:
$ v^2 = u^2 + 2aL $
Since the electron starts from rest, $u = 0$. Therefore, the equation simplifies to:
$ v^2 = 2aL $
Solving for $a$, we get:
$ a = \frac{v^2}{2L} \quad 	ext{...(ii)} $
By equating expressions from (i) and (ii):
$ \frac{qE}{m} = \frac{v^2}{2L} $
Rearranging for $\frac{q}{m}$, we find:
$ \frac{q}{m} = \frac{v^2}{2EL} = \frac{v^2}{2E\ell} \quad (	ext{since } L = \ell) $

Answer

$\frac{\mathrm{v}_2}{2 \mathrm{E} l}$

Answer

$\frac{\mathrm{v}_2 l}{2 \mathrm{E}}$

Answer

$\frac{2 \mathrm{E}}{\mathrm{v}^2 l}$

Answer

$\frac{\mathrm{v}_2 l}{\mathrm{E}}$

Electrostats & Capacitance

An electron of mass ' m ' and charge ' q ' is accelerated from rest in a uniform electric field of intensity ' E '. The velocity acquired by it as it travels a distance ' l ' is ' v '. The ratio \frac{\mathrm{q}}{\mathrm{m}} in terms of E, l and v is

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