Answer:
The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
Explanation:
Given:
(A)
Current
A
Voltage
V
For finding the resistance,



12Ω
(B)
For finding power delivered,


Watt
(C)
For finding the potential difference,



V
Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q =
is located in the center of a spherical cavity of radius ,
m inside an insulating spherical charged solid.
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)


Electric field at
m due to shell
E1 = 

Electric field at
due to 'q' at center 
E2 =

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
![=[ 2.134 - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)

Answer:
8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.
Explanation:
The total charge accumulated in 1.8 hours will be:
Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)
Total Charge = - 12960 nC = - 12.96 x 10^(-6) C
Since, the charge on one electron is e = - 1.6 x 10^(-19) C
Therefore, no. of electrons will be:
No. of electrons = Total Charge/Charge on one electron
No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]
<u>No. of electrons = 8.1 x 10^13 electrons</u>
Answer:
The Arsenic has three electron-containing orbitals. The orbitals s, p and d.
Explanation:
Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:








Therefore, the Arsenic has three electron-containing orbitals (s, p d).