# A physics derivation! HELP!

A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let 'N' be the number density of free electrons, each of mass ′m′. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. lf the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ′ωp', which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ω approaches ωp, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. Taking the electronic charge as 'e' and the permittivity as ′ϵ0', use dimensional analysis to determine the correct expression for ωp.

Note by Ashley Shamidha
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

This is a question from JEE-2011.

The answer is $$\sqrt{\frac{Ne^2}{me_{o}}}$$.

You can see the solution here (question 38).

- 3 years, 3 months ago

Log in to reply

Thank you!

- 3 years, 2 months ago

Log in to reply

You're welcome :)

- 3 years, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...