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3 years ago

Which color has the lowest frequency?

1 answer:

4 0

<h2>RED!</h2><h3></h3><h3>On the visible spectrum, red has the lowest frequency.</h3><h3>(I'm an amateur astronomer, so I would know.)</h3>

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

And so, for r<a,

$V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

- Potential at the surface of the outer cylinder:

$V(b)=0$

Therefore, the potential difference is simply equal to

$V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

The electric field is just the derivative of the electric potential:

$E=-\frac{dV}{dr}$

so we can find it by integrating the expression for the electric potential. We find:

$E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

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7 0

3 years ago

What was created 300 years ago by scientist carolus linnaeus

Norma-Jean [14]

<span>the classification system</span>

6 0

3 years ago

Use the drop-down menus to identify the type of energy each phrase describes.

Tresset [83]

The shortest wavelength of visible light = violet light

Energy that can be felt as heat but not seen = infrared

Short, invisible rays that can cause eye damage = ultraviolet

Visible light with the longest wavelength = red light

Explanation:

Electromagnetic waves are waves consisting of oscillations of the electric and the magnetic field, occurring in a plane perpendicular to the direction of motion the wave.

They are the only type of waves able to travel without a medium, and they are transverse in nature.

All electromagnetic waves travel in a vacuum at the speed of light, which value is:

$c=3.0\cdot 10^8 m/s$

Electromagnetic waves are classified into 7 different classes, depending on their wavelength/frequency, and they have different properties. From shortest to longest wavelength (and from highest to lowest frequency), they are:

Gamma rays

X rays

Ultraviolet

Visible light

Infrared radiation

Microwaves

Radio waves

Moreover, the visible light of the spectrum is further divided into different colors, according to how our eye perceive them; from shortest to longest wavelength:

violet

blue

green

yellow

orange

red

Therefore, we have:

The shortest wavelength of visible light is violet light, which has wavelength between 380 and 450 nm
The longest wavelength of visible light is red light, which has wavelenght between 620 and 750 nm
Infrared radiation is a type of radiation that is felt as heat by our body, however it cannot be seen because it falls outside the spectrum of visible light
Ultraviolet radiation is also invisible to human eye; it has shorter wavelength than visible light and therefore it has more frequency (and more energy), therefore it can cause damage, especially to the eye

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5 0

3 years ago

Read 2 more answers

A type of wave is mechanical - what's the other type?

ANEK [815]

Answer:

There are three types of mechanical waves: transverse waves, longitudinal waves, and surface waves. :/

Explanation: