When compounds form, what is one result for the atoms that bonded?

The deepest section of ocean in the world is the Marianas Trench, located in the Pacific Ocean. Here the ocean floor is as low a

Levart [38]

B I just took the test

3 0

1 year ago

Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

Answer:

For the point inside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

For the point outside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}$

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}$

We have to integrate it over the ring, which is an angular integration.

$E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}$

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

We will only change the boundaries of the last integration.

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

6 0

3 years ago

Many household products we consider necessary today such as AM and FM radios, televisions, wireless networks, cordless and cellu

san4es73 [151]

I want to say frequencies would be the answer. I could be wrong though. Hope this may have helped!

4 0

3 years ago

Read 2 more answers

See the diagram above. Which of the following is the best prediction of what would happen if you increased the distance in secti

Inessa [10]

By the definition of wavelength, the answer is the letter D, the wavelength would decrease.

We can see in the diagram a wave motion.

A wave has some characteristics:

Has an amplitude, the distance from 0 to the crest (highest point in the y-direction, point (3) in the figure) it would see in the figure as (2)
Has wavelength, the distance between the crests.
Has a trough, the lowest point in the y-direction.

Now, if we increase the distance of the crests, by the definition shown above, we will increase the wavelength.

Therefore, the answer is letter D, the wavelength would increase.

You can learn more about wave motion here:

brainly.com/question/22763521

7 0

2 years ago

What would happened if there is no invented of machine?

yuradex [85]

if there were no invention of machines then life would have been more difficult and simple works could be hard to do. Even now we are using our phones, sitting in a AC room interacting to eachother from different places. without the invention of machines simple things like transportation would have been difficult. There would be horses and donkey for the transportation. There would be no electricity,no internet, no transportation, not even c computers or mobile etc. The market for business will be smaller, the knowledge and news about world would be less.

so the problem would have been bigger than we can imagine. But one thing is that nature could survive lot more compared to what we have done till now by destroying nature.

4 0

1 year ago