A straight, cylindrical wire lying along the x axis has a length L and a diameter d . It is made of a material described by Ohm'
1 answer:
The magnitude and direction of the electric field in the wire are mathematically given as
Generally, the equation for is mathematically given as
A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides
E d
In conclusion, the magnitude and direction of the electric field in the wire are given as
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1) In the reference frame of one electron: 0.38c
To find the relative velocity of one electron with respect to the other, we must use the following formula:
where
u is the velocity of one electron
v is the velocity of the second electron
c is the speed of light
In this problem:
u = 0.2c
v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)
Substituting, we find:
2) In the reference frame of the laboratory: -0.2c and +0.2c
In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are
+0.2c
-0.2c
1) Linear charge density of the shell:
2) x-component of the electric field at r = 8.7 cm: outward
3) y-component of the electric field at r =8.7 cm: 0
4) x-component of the electric field at r = 1.15 cm: outward
5) y-component of the electric field at r = 1.15 cm: 0
Explanation:
1)
The linear charge density of the cylindrical insulating shell can be found by using
where
is charge volumetric density
A is the area of the cylindrical shell, which can be written as
where
is the outer radius
is the inner radius
Therefore, we have :
2)
Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.
The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:
where:
where
is the linear charge density of the wire
r = 8.7 cm = 0.087 m is the distance from the axis
And this field points radially outward, since the charge is positive .
And
where
And this field points radially inward, because the charge is negative.
Therefore, the net field is
in the outward direction.
3)
To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.
However, we notice that since the wire is infinite, for the element of electric field produced by a certain amount of charge
along the wire there exist always another piece of charge
on the opposite side of the wire that produce an element of electric field
, equal and opposite to
.
Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.
We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,
4)
Here we want to find the x-component of the electric field at a point at
r = 1.15 cm
from the central axis.
We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.
Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:
where:
is the linear charge density of the wire
r = 1.15 cm = 0.0115 m is the distance from the axis
This field points radially outward, since the charge is positive . Therefore,
5)
For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field produced by a certain amount of charge
along the wire there exist always another piece of charge
on the opposite side of the wire that produce an element of electric field
, equal and opposite to
.
Therefore, the y-component of the electric field is zero.
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