At the moment $t = 0$ , an electron leaves one plate of a parallel-plate condenser with a negligible velocity. An accelerating voltage varying as $V = a t$ , where $a$ is a constant is applied between the plates. The separation between the plates is $l$ . The velocity of the electron at the moment it reaches the opposite plate will be :

{(\frac{2 e a l}{9 m})}^{\frac{1}{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(\frac{3 e a l}{4 m})}^{\frac{1}{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(\frac{4 e a l}{m})}^{\frac{1}{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(\frac{9 e a l}{2 m})}^{\frac{1}{3}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D ${(\frac{9 e a l}{2 m})}^{\frac{1}{3}}$
The accelerating voltage varies as $V = a t$ .
Acceleration of the electron at $t$ , $α = \frac{e E}{m}$
But electric field between the plates at time $t$ , $E = \frac{V}{l} = \frac{a t}{l}$
We know, $α = \frac{d v}{d t}$
$∴$ $d v = \frac{e a t}{m l} d t$
Integrating, we get $\int_{o}^{v (t)} = \frac{e a}{m l} \int_{o}^{t} t d t$
$⟹$ $v (t) = \frac{e a t^{2}}{2 m l}$
Now $d x = v (t) d t$
$∴$ $\int_{o}^{l} d x = \frac{e a}{2 m l} \int_{0}^{T} t^{2} d t$
$⟹$ $l = \frac{e a T^{3}}{6 m l}$
We get, time after which the electron reaches the opposite plate $T = (\frac{6 m l^{2}}{e a})^{1 / 3}$
So, velocity of electron after reaching opposite plate is
$v (T) = \frac{e a}{2 m l}$ $(\frac{6 m l^{2}}{e a})^{2 / 3} =$ $(\frac{9 e a l}{2 m})^{1 / 3}$

Suggest Corrections

Similar questions

A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of $2 \times 10^{6} m s^{- 1}$ moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?