Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

An electron moving in a gravitational free space enters a uniform electric field $E$ with an initial velocity $v_{0}$ as shown in the figure.

$(a)$ Find the deflection distance undefined in the field.

$(b)$ Find an expression for the velocity of the electron when it just emerges from the field.

$(c)$ Find an expression for the total deflection distance $H$ at a vertical screen placed at a distance $l$ from the region of the uniform field. (Assume that the field abruptly ends outside the field.)

Accepted Answer

Given that the total deflection distance, $H$

Distance of screen from the region of the uniform field, $l$

The initial velocity of the electron, $v_{0}$

Time taken by the electron to cross the field, $t = \frac{l}{v_{0}}$

Displacement of the particle outside the field, $y'$

When the electron moves out of the electric field, it moves in a straight line with its final velocity.

Hence, we can apply the law of motion.

Deflection of the electron in the field is the vertical distance covered by the electron in the field with the vertical acceleration $a$ .

$h = u t + \frac{1}{2} a t^{2}$

$h = 0 + \frac{1}{2} (\frac{e E}{m}) {(\frac{l}{v_{0}})}^{2} = (\frac{e E}{2 m v_{0}^{2}}) l^{2}$

where $m$ is mass and $e$ is the charge on the electron.

The velocity of the electron when it comes out of the plate is

$\vec{v} = \vec{u} + \vec{a t} = v_{0} \hat{i} + (\frac{e E}{m}) t \hat{j}$

$E$

Vertical displacement of the particle outside the field, $y'$

When the electron moves out of the electric field, it moves in a straight line with its final velocity.

$y' = (v_{y}) \cdot t'$ ,

where $v_{y}$ is the vertical velocity of electron outside the field and $t'$ is time taken by the electron to reach screen after leaving the field.

$y' = (\frac{e E l}{m v_{0}}) (\frac{l}{v_{0}}) = \frac{e E l^{2}}{m v_{0}^{2}}$

Hence, $H = h + y' = \frac{e E l^{2}}{2 m v_{0}^{2}} + \frac{e E l^{2}}{m v_{0}^{2}} = \frac{3 e E l^{2}}{2 m v_{0}^{2}}$ .

This is the required expression for $H$ .

Important Questions on Coulomb's Law and Electric Field

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