Answer:
The surface charge density on planes A and B respectively is
![\sigma__{A}} } = 2\sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%202%5Csigma)
and
![\sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BB%7D%7D%20%3D%20%20%20%5Csigma)
Explanation:
From the question we are told that
The electric field in region to the left of A is ![E_i = \frac{3 \sigma}{2 \epsilon_o}](https://tex.z-dn.net/?f=E_i%20%3D%20%20%5Cfrac%7B3%20%5Csigma%7D%7B2%20%5Cepsilon_o%7D)
The direction of the electric field is left
The electric field in the region to the right of B is ![E_f = \frac{3 \sigma}{2 \epsilon_o}](https://tex.z-dn.net/?f=E_f%20%3D%20%20%5Cfrac%7B3%20%5Csigma%7D%7B2%20%5Cepsilon_o%7D)
The direction of the electric field is right
The electric field in the region between the two planes is ![E_m = \frac{\sigma }{2 \epsilon_o }](https://tex.z-dn.net/?f=E_m%20%20%3D%20%20%5Cfrac%7B%5Csigma%20%7D%7B2%20%5Cepsilon_o%20%7D)
The direction of the electric field is right
Let the surface charge density on planes A and B be represented as ![\sigma__{A}} \ \ and \ \ \sigma__{B}} \ \ \ respectively](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%20%5C%20%5C%20and%20%20%5C%20%5C%20%5Csigma__%7BB%7D%7D%20%5C%20%5C%20%5C%20%20respectively)
From the question we see that
![E_i = E_f](https://tex.z-dn.net/?f=E_i%20%3D%20E_f)
Generally the electric to the right and to the left is due to the combined electric field generated by plane A and B so
![E_i = E_f = \frac{3\sigma }{2\epsilon} = \frac{\sigma_A }{ 2 \epsilon_o } + \frac{\sigma_B }{ 2 \epsilon_o }](https://tex.z-dn.net/?f=E_i%20%3D%20E_f%20%3D%20%20%5Cfrac%7B3%5Csigma%20%7D%7B2%5Cepsilon%7D%20%20%3D%20%20%5Cfrac%7B%5Csigma_A%20%7D%7B%202%20%5Cepsilon_o%20%7D%20%2B%20%20%5Cfrac%7B%5Csigma_B%20%7D%7B%202%20%5Cepsilon_o%20%7D)
=> ![\sigma__{A}} + \sigma__{B}} = 3 \sigma -- -(1)](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%2B%20%5Csigma__%7BB%7D%7D%20%3D%20%203%20%5Csigma%20%20--%20-%281%29)
Generally the electric field at the middle of the plane A and B is due to the diffencence in electric field generated by plane A and B
i.e
![\frac{\sigma }{2 \epsilon_o } = \frac{\sigma_A }{ 2 \epsilon_o } - \frac{\sigma_B }{ 2 \epsilon_o }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%20%7D%7B2%20%5Cepsilon_o%20%7D%20%20%3D%20%20%5Cfrac%7B%5Csigma_A%20%7D%7B%202%20%5Cepsilon_o%20%7D%20-%20%20%20%5Cfrac%7B%5Csigma_B%20%7D%7B%202%20%5Cepsilon_o%20%7D)
=> ![\sigma__{A}} - \sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20-%20%20%5Csigma__%7BB%7D%7D%20%3D%20%20%5Csigma)
=> ![\sigma__{A}} } = \sigma + \sigma__{B}](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%20%5Csigma%20%2B%20%5Csigma__%7BB%7D)
From equation 1
![\sigma + \sigma__{B}}+ \sigma__{B}} = 3 \sigma](https://tex.z-dn.net/?f=%5Csigma%20%2B%20%5Csigma__%7BB%7D%7D%2B%20%5Csigma__%7BB%7D%7D%20%3D%20%203%20%5Csigma)
=> ![\sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BB%7D%7D%20%3D%20%20%20%5Csigma)
So
![\sigma__{A}} } = \sigma + \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%20%5Csigma%20%2B%20%5Csigma)
=> ![\sigma__{A}} } = 2\sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%202%5Csigma)
Answer:
Explanation: heat of fusion
Answer:
The gravitational potential energy of the skydiver=![79200 k g m^{2} / s^{2}](https://tex.z-dn.net/?f=79200%20k%20g%20m%5E%7B2%7D%20%2F%20s%5E%7B2%7D)
Given:
Mass of the skydiver=80kg
Distance of the skydiver from the earth=100m
To find:
Gravitational potential energy of the skydiver
<u>Step by Step Explanation:</u>
Solution:
According to the formula, Gravitational Potential Energy (GHE) is calculated as
Where m= Mass of the skydiver
g=Acceleration due to gravity
h=Distance of the skydiver from the earth
Substitute the known values in the above equation we get
![G H E=80 \times 9.8 \times 100](https://tex.z-dn.net/?f=G%20H%20E%3D80%20%5Ctimes%209.8%20%5Ctimes%20100)
![=79200 \mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}](https://tex.z-dn.net/?f=%3D79200%20%5Cmathrm%7Bkg%7D%20%5Cmathrm%7Bm%7D%5E%7B2%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D)
Result:
Thus the gravitational potential energy of the skydiver is =![79200 k g m^{2} / s^{2}](https://tex.z-dn.net/?f=79200%20k%20g%20m%5E%7B2%7D%20%2F%20s%5E%7B2%7D)
Answer: The speed of the student is 5.45km/hr.
Explanation: The important concept here is to convert 33 minutes to hours by diving by 60 since one hour is equal to 60 minutes to get 0.55hr
Then velocity is equal to distance over time.
v=d/t
v= 3km/0.55hr
v=5.45km/hr