What is a photovoltaic cell
1 answer:
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a) E = 0
b)
Explanation:
a)
We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:
where
E is the electric field
q is the charge contained by the Gaussian surface
is the vacuum permittivity
Here we want to find the electric field at a distance of
r = 12 cm = 0.12 m
Here we are between the inner radius and the outer radius of the shell:
However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.
This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:
q = 0
Therefore, the magnitude of the electric field is also zero:
E = 0
b)
Here we want to find the magnitude of the electric field at a distance of
r = 20 cm = 0.20 m
from the centre of the shell.
Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.
Therefore, it is given by:
where in this problem:
is the charge on the shell
is the distance from the centre of the shell
Substituting, we find:
Answer:
The temperature of the core raises by every second.
Explanation:
Since the average specific heat of the reactor core is 0.3349 kJ/kgC
It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius
Thus reactor core whose mass is will require
energy to raise it's temperature by 1 degree Celsius in 1 second
Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by