The highest energy waves have the

A very long, uniformly charged cylinder has radius R and linear charge density λ. Find the cylinder's electric field strength ou

mixer [17]

The cylinder's electric field magnitude, at a distance <em>r</em> from the axis of the cylinder (greater than the cylinder's radius), is equal to $E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

<h3>Further explanation</h3>

Matter is the building block of everything that we encounter in our lives. Matter is made of atoms, which are in turn made of tiny particles which are called electrons, protons, and neutrons. The ammount of these 3 elements, and their topological configuration in the atoms, is what determines what a certain element is (like Carbon, Hydrogen, Iron, etc).

In some cases, some elements may lose or gain some electrons. Regarded that this missing or extra electrons are not very high in number, the material doesn't lose any of its properties, however it will always try to get its number of electrons back to normal. This is when we say that an element has a <em>charge</em>, which is a measure of how much electrons a body needs to get back to normal. A body has positive charge if it lacks electrons, and has negative charge if it has extra electrons.

This charge causes the material to have an Electric field, which is a measure of how much does it attract or repel electrons. In the case of our problem, we need to compute exactly that, the Electric field. In our problem, we have an infinitely long cylinder with a linear charge density $\lambda$ , this means that all parts of the cylinder have the same charge, and due to symmetry, the electric field is constant on the angular and longitudinal directions of the cylinder.

This makes easy to apply Gauss' Law, since for a Gaussian curve in the shape of a concentric cylinder (with a higher radius than that of our charged cylinder) we can write:

$\Phi = \frac{\lambda \cdot L}{\epsilon_0}$

Where $\Phi$ is called the Electric flux. Since the electric field is constant for a given distance <em>r</em> from the axis of the cylinder we can write that:

$\Phi = E \cdot 2\pi r \cdot L$

Joining both our expressions we can get that:

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

<h3 /><h3>Learn more</h3>

Description on Electric fields: brainly.com/question/8971780
Relation between electric fields and magnetism: brainly.com/question/2838625
How can we use electric charges: brainly.com/question/10427437

<h3>Keywords</h3>

Electrons, protons, electric field, cylinder, electric flux

5 0

3 years ago

Read 2 more answers

What current is used to power the United States power grid?

alexdok [17]

The answer is Alternating Current

3 0

2 years ago

Read 2 more answers

Would water molecules in Venus’ atmosphere, whose temperature is 740 K, escape into outer space? A water molecule has a mass tha

Akimi4 [234]

Answer:

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.

Explanation:

The average speed of gas molecules is given by:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

R is the gas constant, T is the temperature and M the molar mass of the gas.

We know that a water molecule has a mass that is 18 times that of a hydrogen atom:

$M_H=1.01*10^{-3}\frac{kg}{mol}\\M_{H2O}=18M_H=0.02\frac{kg}{mol}$

So, we have:

$v_{rms}=\sqrt{\frac{3(8.314\frac{J}{mol \cdot K})740K}{0.02\frac{kg}{mol}}}\\v_{rms}=960.65\frac{m}{s}*\frac{1km}{1000m}=0.96\frac{km}{s}$

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:

$10\frac{km}{s}*\frac{1}{6}=1.6\frac{km}{s}\\0.96\frac{km}{s}$

4 0

3 years ago

Most earthquakes occur along or near the edges of the earth's

Pepsi [2]

Most earthquakes occur along or near the edges of the earth's lithospheric<span> plate. </span>

8 0

3 years ago

A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How

nalin [4]

Answer:

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

6 0

3 years ago

The highest energy waves have the _ frequency