Ask question

Ask question

All categories

2 years ago

5

Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons a

re moved from one plate to the other, what is the electric field between the plates?

1 answer:

dsp732 years ago

8 0

Answer:

$E=576.5V/m$

Explanation:

From the question we are told that:

Length $l=56.0cm=0.56m$

Distance apart $d=7.0mm=0.007m$

Electron Transferred $n=10^{-10}$

Therefore

Total Charge

Since Charge on each electron is

$e=1.602*10^{-19}$

Therefore

$T=1.602*10^{-19} *10^{10}$

$T=1.602*10^{-9}$

Generally the equation for Charge density is mathematically given by

$\rho=T/A$

Where

Area

$A=0.56*0.56$

$A=0.3136$

Therefore

$\rho=1.602*10^{-9}/0.3136$

$\rho=5.10*10^{-9}$

Generally the equation for Electric Field in the capacitor is mathematically given by

$E=\frac{\rho}{e_0}$

$E=\frac{5.10*10^{-9}}{8.85x10{-12}}$

$E=576.5V/m$

You might be interested in

Mrs. Miller is acting a little strange and is pushing as hard as she can against a wall. Using Newton's Laws, what could be say

deff fn [24]

Newton’s 3rd Law of Motion

8 0

2 years ago

Read 2 more answers

How many grams of co2 are in 3 moles of the compound

Ber [7]

Answer:132.0285

Explanation: Hope this helps!

7 0

2 years ago

Read 2 more answers

Help pls i’m not sure what to do

Sav [38]

Answer:

I DONT KNOW WHAT TO DO SORRY

Explanation:

EVEN ME IM NOT SURW

7 0

2 years ago

Which of the following is not true of a form?

Ludmilka [50]

Answer:

It can only display one record at a time

Explanation:

Form ;

1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.

2.This is the most popular method of data entry

3.It may contain images in the background.

4.This can be sorted data regardless of its source of information.

Only option C is wrong.

Therefore the answer C is correct.

4 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago