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

VikaD [51]

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.

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.

We will only change the boundaries of the last integration.

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

6 0

3 years ago

Which circuit would have the most electrical power?

aliya0001 [1]

the answer is c because it has the most volts

a

6 0

2 years ago

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as , where a

Misha Larkins [42]

Answer:

3 times louder

Explanation:

The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².

Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹

and I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₁ = 10㏒(I₁/I₀)

L₁ = 10㏒(10⁻⁹/10⁻¹²)

L₁ = 10㏒(10³)

L₁ = 3 × 10㏒10

L₁ = 30㏒10

L₁ = 30 dB

Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₂ = 10㏒(I₁/I₀)

L₂ = 10㏒(10⁻³/10⁻¹²)

L₂ = 10㏒(10⁹)

L₂ = 9 × 10㏒10

L₂ =90㏒10

L₂ = 90 dB

So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3

So, Braylee's music is 3 times louder than Jessica's music

6 0

3 years ago

Explain how the thermal energy of an isolated system changes with time if the mechanical energy of that system is constant.

Naddika [18.5K]

Answer:

Thermal energy of an isolated system changes with time If the mechanical energy of that system is constant according to the first law of thermodynamics, which states that thermal energy of an isolated system can still change as long as the total energy of that system does not change.

Explanation:

6 0

2 years ago

Đổi 20độC = ? độ F =? Độ k

san4es73 [151]

Answer:

IDC

Explanation:

I DON'T UNDERSTAND........

7 0

2 years ago