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PilotLPTM [1.2K]

1 year ago

9

when charges mutually repel and distribute themselves on the surface of conductors, what becomes of the electric field inside th

e conductor?

1 answer:

AleksAgata [21]1 year ago

3 0

The charges align themselves so that the conductor's internal field is zero.

<h3>What occurs if a charged surface is in close proximity to a conducting surface?</h3>

Induced charges are created on the conductor when a charge is brought close to it. The internal free charges of the conductor, however, are gathered throughout its surface because the electric field inside the conductor must be zero in order to defeat the electric field of induced charges.

<h3>What takes place within a conductor?</h3>

A substance that has a lot of free electrons accessible for the flow of current is said to be a conductor. Since there are numerous electrons, a powerful force of repulsion exists between them as well. As a result, the electrons move to lessen their attraction for one another.

To know more about conductor's visit:-

brainly.com/question/18084972

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A 1.5m wire carries a 7 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

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A torus is formed when a circle of radius 3 centered at (5 comma 0 )is revolved about the y-axis. a. Use the shell method to wr

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Answer:

a) $V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx$

b) $V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy$

c) $V=90\pi ^{2}$

Explanation

In order to solve these problems, we must start by sketching a drawing of what the graph of the problem looks like, this will help us analyze the drawing better and take have a better understanding of the problem (see attached pictures).

a)

On part A we must build an integral for the volume of the torus by using the shell method. The shell method formula looks like this:

$V=\int\limits^a_b {2\pi r y } \, dr$

Where r is the radius of the shell, y is the height of the shell and dr is the width of the wall of the shell.

So in this case, r=x so dr=dx.

y is given by the equation of the circle of radius 3 centered at (5,0) which is:

$(x-5)^{2}+y^{2}=9$

when solving for y we get that:

$y=\sqrt{9-(x-5)^{2}}$

we can now plug all these values into the shell method formula, so we get:

$V=\int\limits^8_2 {2\pi x \sqrt{9-(x-5)^{2}} } \, dx$

now there is a twist to this problem since that will be the formula for half a torus.Luckily for us the circle is symmetric about the x-axis, so we can just multiply this integral by 2 to get the whole volume of the torus, so the whole integral is:

$V=\int\limits^8_2 {4\pi x \sqrt{9-(x-5)^{2}} } \, dx$

we can take the constants out of the integral sign so we get the final answer to be:

$V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx$

b)

Now we need to build an integral equation of the torus by using the washer method. In this case the formula for the washer method looks like this:

$V=\int\limits^b_a{\pi(R^{2}-r^{2})} \, dy$

where R is the outer radius of the washer and r is the inner radius of the washer and dy is the width of the washer.

In this case both R and r are given by the x-equation of the circle. We start with the equation of the circle:

$(x-5)^{2}+y^{2}=9$

when solving for x we get that:

$x=\sqrt{9-y^{2}}+5$

the same thing happens here, the square root can either give you a positive or a negative value, so that will determine the difference between R and r, so we get that:

$R=\sqrt{9-y^{2}}+5$

and

$r=-\sqrt{9-y^{2}}+5$

we can now plug these into the volume formula:

$V=\pi \int\limits^3_{-3}{(5+\sqrt{9-y^{2}})^{2}-(5-\sqrt{9-y^{2}})^{2}} \, dy$

This can be simplified by expanding the perfect squares and when eliminating like terms we end up with:

$V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy$

c) We are going to solve the integral we got by using the washer method for it to be easier for us to solve, so let's take the integral:

$V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy$

This integral can be solved by using trigonometric substitution so first we set:

$y=3 sin \theta$

which means that:

$dy=3 cos \theta d\theta$

from this, we also know that:

$\theta=sin^{-1}(\frac{y}{3})$

so we can set the new limits of integration to be:

$\theta_{1}=sin^{-1}(\frac{-3}{3})$

$\theta_{1}=-\frac{\pi}{2}$

and

$\theta_{2}=sin^{-1}(\frac{3}{3})$

$\theta_{2}=\frac{\pi}{2}$

so we can rewrite our integral:

$V=20\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\sqrt{9-(3 sin \theta)^{2}}} \, 3 cos \theta d\theta$

which simplifies to:

$V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-(3 sin \theta)^{2}}} \, cos \theta d\theta$

we can further simplify this integral like this:

$V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-9 sin^{2} \theta}}} \, cos \theta d\theta$

$V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {3(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{cos^{2} \theta})}} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(cos \theta})} \, cos \theta d\theta$

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {cos^{2} \theta}} \, d\theta$

We can use trigonometric identities to simplify this so we get:

$V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\frac{1+cos 2\theta}{2}}} \, d\theta$

$V=90\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {1+cos 2\theta}}} \, d\theta$

we can solve this by using u-substitution so we get:

$u=2\theta$

$du=2d\theta$

and:

$u_{1}=2(-\frac{\pi}{2})=-\pi$

$u_{2}=2(\frac{\pi}{2})=\pi$

so when substituting we get that:

$V=45\pi\int\limits^{\pi}_{-\pi} {1+cos u}} \, du$

when integrating we get that:

$V=45\pi(u+sin u)\limit^{\pi}_{-\pi}$

when evaluating we get that:

$V=45\pi[(\pi+0)-(-\pi+0)]$

which yields:

$V=90\pi ^{2}$

8 0

3 years ago