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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

A 800hz tuning fork is vibrating, producing a sound wave in the air.

blsea [12.9K]

Answer:

The frequency of the sound wave is 800Hz

The speed of sound in a is about 340m/s.

Velocity = frequency x wavelength

making wavelength the subject formula

wavelength = Velocity/frequency.

wavelength = 340/800

wavelength = 0.425m.

3 0

2 years ago

A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is

puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

$\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}$

$\dfrac{1}{u}=\dfrac{103}{4080}$

$u =\dfrac{4080}{103}$

The magnification is

$m = \dfrac{-v}{u}$

$m=\dfrac{-240\times103}{4080}$

$m = -6.05$

Hence, The magnification is -6.05.

6 0

3 years ago

In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles

alexandr402 [8]

Answer:

Explanation:

We shall apply conservation of mechanical energy

kinetic energy of alpha particle is converted into electric potential energy.

1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³

= 484.4 x 10⁻¹⁶

=48.4 x 10⁻¹⁵ m

8 0

4 years ago

A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnit

nordsb [41]

Answer:

D

Explanation:

Ball A is a non positively charged non metal while ball B is metal ball.

Given: The ball B positive charge of small magnitude

To prove: Balls will attract each other

IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

Hence the correct answer will be D .

8 0

3 years ago