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An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ 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: $1.28\cdot 10^7 N/C$ 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

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

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:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ 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

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

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 $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

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,

$E_y=0$

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:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ 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,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

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 $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

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Lolilolololilolollololililili

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Which of the following are true statements about the electric field? a. A moving charge produces a magnet anomaly. b. The electr

mars1129 [50]

Answer:

a,b,c,d,,f, g, j

Explanation:

e) equipotential lines are lines connecting points of equal potential

h) electric field inside the conductor is non-zero even when there is net movement of charge or non-zero current.

i) capacitors' plates are charged and an electric field exists between the plates.

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A fishing boat accidentally spills 3.0 barrels of diesel oil into the ocean. each barrel contains 42 gallons. if the oil film on

9966 [12]

Number of barrels are 3.0. Each barrel contains 42 gallons of oil. Thus, total volume of oil will be 42×3=126 gallons.

Converting gallons into m^{3}

1 gallon=0.00378 m^{3}

Thus, 126 gallons=0.4769 m^{3}

Thickness of oil film is 2.5\times 10^{2} nm, converting it into meters as follows:

1 nm=10^{-9} m

Thus,

2.5\times 10^{2} nm=1.5\times 10^{-7}m

Now, volume V of oil is related to area A and thickness T as follows:

V=A×T

rearranging,

A=\frac{V}{T}=\frac{0.4769 m^{3}}{2\times 10^{-7}m}=2.38\times 10^{6}m^{2}

Thus, square meters of oil will be 2.38\times 10^{6}m^{2}

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Spring is here, and Cho and her uncle would like to go fishing for the weekend in Washington. Cho could either go to the river i

aleksley [76]

Answer:

B. True since when a person removes the weights there is less for the other people

Explanation:

This problem ask us consider some things

In the private current it is considered a good example of a private good

the fish in the river are a common resource

the fishing in the river is tragedy of the commons

A False. The fish in the river are natural and not sown

B. True since when a person removes the weights there is less for the other people

C. False. Some fishermen will go to the creek

D. False some will go to the river

.

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Pls help me quickly ......​

Pls help me quickly ......