1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Step2247 [10]
3 years ago
9

Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

n the axis of symmetry, both inside and outside of the cylinder, by using the field superposition method. For example, break the cylinder into infinitesimally wide rings, and integrate down the length of the cylinder or break the cylinder into infinitesimally wide, semi-infinite lines, and integrate around the circumference of the cylinder (you can use symmetry to simplify this calculation).
Physics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

For the point inside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

For the point outside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}

We have to integrate it over the ring, which is an angular integration.

E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

<u>x = +x0: </u>

We will only change the boundaries of the last integration.

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

You might be interested in
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
How does the force of gravity between two bodies change when the distance between them doubles? 1. unable to determine; the mass
Rzqust [24]
6. Drop to one quarter of its original value
7 0
3 years ago
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

\theta_{max} =18.38^o

b

New  n_{cladding} =1.491

Explanation:

 From the question we are told that

          The refractive index of the core is  n_{core} = 1.497

         The refractive index of the cladding  is   n_{cladding} = 1.421

Generally according to Snell's law

      n_{core} * sin(90- \theta) = n_{cladding} * sin (90)

Where \theta_{max} is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection

      \theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]

       \theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]

      \theta_{max} =18.38^o

Given from the question the the largest angle is  5°

Generally the refraction index of the cladding is mathematically represented as

           n_{cladding} = n_{core} * sin (90 - 5)

          n_{cladding} =1.491

       

5 0
2 years ago
Trong thí nghiệm về giao thoa sóng trên mặt nước gồm hai nguồn kết hợp S1S2 cách nhau 15 cm với dao động với tần số 30Hz. Tốc độ
mojhsa [17]

Answer:

this is a difficult question but I will try to answer it answer for this is 3220 a + b b u s y d l new

6 0
2 years ago
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l
butalik [34]

Answer:

a)   f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} } , b)   Δf = 2 f₀ \frac{v}{c}

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo\frac{c+v}{c}

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ \frac{c}{ c-v}

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   \frac{c}{c-v}

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} }

         

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 ( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}

 

                f ’’ = fo ( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )

                f ’’ = fo ( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )

leave the linear term

               f ’’ = f₀ + f₀ 2\frac{v}{c}

the sound difference

               f ’’ -f₀ = 2f₀ v/c

               Δf = 2 f₀ \frac{v}{c}

4 0
2 years ago
Other questions:
  • What is the movement of alternating compressions and rarefactions?
    9·1 answer
  • A train travels 98 kilometers in 1 hours, and then 78 kilometers in 3 hours. what is its average speed?
    14·1 answer
  • You and your friends are going on a picnic in a very small car. On your way, the car breaks down and you try to push it to get i
    9·1 answer
  • How do you make a tissue dance
    9·1 answer
  • A galvanometer can be calibrated to read electric A) current B) Voltage C) either of these D) neither of these
    8·1 answer
  • A bicycle has a mass of 10kg and an acceleration of 2m/s². What is the net force of the bicycle?
    14·1 answer
  • An infinitely long line charge of uniform linear charge density λ = -3.00 µC/m lies parallel to the y axis at x = -3.00 m. A poi
    10·1 answer
  • A capacitor consists of two parallel conducting plates, each of area 0.4 m2 and separated by a distance of 2.0 cm. Assume there
    5·1 answer
  • An apple contains 165 Calories. How many actual calories does it contain? How many joules does it contain
    6·1 answer
  • What happens to the heat energy when you increase the length of an object​
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!