Choose the correct statement concerning units of power or energy. a. A kilojoule (kJ) is a unit of power. b. A gigawatt (GW) is
1 answer:
Answer:
A kilowatt (kW) is a unit of power.
Explanation:
The power of an object is given by :
Here,
E is the energy required
t is time
The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.
Option (1) : A kilojoule (kJ) is a unit of power is incorrect.
Option (2) : A gigawatt (GW) is a unit of energy is incorrect.
Option (3) : A watt (W) is a unit of energy is incorrect.
Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.
Option (4) : A kilowatt (kW) is a unit of power is correct.
Hence, the correct option is (d).
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a)
i) Potential for r < a:
ii) Potential for a < r < b:
iii) Potential for r > b:
b) Potential difference between the two cylinders:
c) Electric field between the two cylinders:
Explanation:
a)
Here we want to calculate the potential for r < a.
Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is
where
is the linear charge density
r is the distance from the wire/surface of the cylinder
By integration, we find an expression for the electric potential at a distance of r:
Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:
So the potential where the electric field is zero is constant:
iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density and an equal negative charge density
. Therefore, the net charge is zero, so the electric field is zero.
This means that the electric potential is constant, so we can write:
However, we know that the potential at b is zero, so
ii) The electric field in the region a < r < b instead it is given only by the positive charge distributed over the surface of the inner cylinder of radius a, therefore it is
And so the potential in this region is given by:
(1)
i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):
E = 0
This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):
And so, for r<a,
b)
Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.
We have:
- Potential at the surface of the inner cylinder:
- Potential at the surface of the outer cylinder:
Therefore, the potential difference is simply equal to
c)
Here we want to find the magnitude of the electric field between the two cylinders.
The expression for the electric potential between the cylinders is
The electric field is just the derivative of the electric potential:
so we can find it by integrating the expression for the electric potential. We find:
So, this is the expression of the electric field between the two cylinders.
Learn more about electric fields:
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