Im reading a book with 398 pages. use p to represent the number of pages i used

Consider a long cylindrical charge distribution of radius R with a uniform charge density rho. Find the electric field at distan

ahrayia [7]

Answer: Ok, first lest see out problem.

It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.

Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0

where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.

So Q= rho*volume= pi*r*r*L*rho

so replacing : E = (1/2)*r*rho/ε0

you may ask, ¿why dont use R on the solution?

since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.

R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.

3 0

3 years ago

Describe one way in which light acts like a wave and one way in which light acts like a particle.

monitta

Answer:

Quantum mechanics tells us that light can behave simultaneously as a particle or a wave. ... When UV light hits a metal surface, it causes an emission of electrons. Albert Einstein explained this "photoelectric" effect by proposing that light – thought to only be a wave – is also a stream of particles.

Explanation:

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3 0

2 years ago

A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the

GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

$E=\phi + K$ (1)

where

$E=\frac{hc}{\lambda}$ is the energy of the incident photon, with

h = Planck constant

c = speed of light

$\lambda$ = wavelength

$\phi$ = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

$eV=K$

So we can rewrite (1) as

$E=\phi + eV$

where we have:

$\lambda=200 nm = 2\cdot 10^{-7} m$

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J$

Converting into electronvolts,

$E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV$

And now we can solve eq.(1) to find the work function of the metal:

$\phi = E-eV=6.22 eV-1.93 eV=4.29 eV$

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so $\phi$ is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

6 0

3 years ago

Why does the cannon move backward when the cannon ball is shot?

seropon [69]

Answer:

This came to mind

Explanation:

when a cannon fires (in real life or in the movies) have noticed that the cannon recoils, sliding backwards after the explosion. Again, a non-zero net force on the cannon changes its momentum.

6 0

2 years ago

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2. Compute the wavelength of this line for l H a

irina1246 [14]

Answer:

$\lambda=550\ nm$