Question

A charged particle of mass m is released from rest x along electric field e.j^ (vector) find angular momentum of particle from origin

Answer 1

Answer on Question #44135, Physics, Electrodynamics

A charged particle of mass $m$ is released from rest $x$ along electric field $e.j^{\wedge}$ (vector) find angular momentum of particle from origin

Solution :

Distinguish three axes $x, y, z$ they correspond to the three vectors $x: \bar{i}, y: \bar{j}, z: \bar{k}$

Then particle has $\{x,0,0\}$ coordinates, and field $\bar{E} = \{0,e,0\}$

Impulse (momentum) from Newton's second law :

$$\frac{d \vec{p}}{dt} = \bar{F} = q \bar{E} = q * \bar{j} * e$$

Obvious that $\vec{p} = qet * \bar{j}$

From angular momentum (L) definition:

$$\bar{L} = [\bar{r} * \vec{p}] = (x * \bar{i} + y * \bar{j} + z * \bar{k}) * (qet * \bar{j}) \text{ (cross vector product)}$$

But $z$ always equal to zero, because there is no force in $z$ direction.

$$\bar{L} = (x * \bar{i} + y * \bar{j}) * (qet * \bar{j}) = qet * (x * \bar{i} + y \bar{j}) * \bar{j} = qet x * \bar{k}$$

$\bar{j} * \bar{j} = 0$ (cross vector product)

https://www.AssignmentExpert.com