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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?
Physics
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

#SPJ4

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In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to
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2.837% less than actual value.

Explanation:

Based on given information let's calculate our value.

S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.

1mile = 1609.34meters.

percentage error is.

\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%

negative indicates less than actual value.

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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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Working...

length of wire L = 1.5 m

current I = 7 A

potential difference V = 68 Volt

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R = V/I

R = 68/7

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Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea
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33,02 lb

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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
arlik [135]

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
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