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Vesnalui [34]
3 years ago
10

A net charge is placed on a hollow conducting sphere. How does the net charge distribute itself? A net charge is placed on a hol

low conducting sphere. How does the net charge distribute itself? The net charge uniformly distributes itself on the sphere's inner and outer surfaces. The net charge clumps together at some location within the sphere. The net charge uniformly distributes itself on the sphere's outer surface. The net charge uniformly distributes itself throughout the thickness of the conducting sphere. The net charge uniformly distributes itself on the sphere's inner surface.
Physics
1 answer:
makvit [3.9K]3 years ago
5 0

Answer:

The net charge uniformly distributes itself on the sphere's inner and outer surfaces.

Explanation:

In electrostatic condition The charge will be distributed over the surface  of the shell,

The electic field outside the sphere  is like the charge where concentrated in the center of the sphere, inside the sphere the electric field is cero (quivalent to say the potencial is constant, non  electric current flow) If the thickness of the shell is very very thin we just have a charged layer. To keep the above mentioned electrostatic condition the elecrtic field have to be cero in this region  too

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The unit of work is called derived unit.Why​
Artemon [7]

Answer:

the unit of work is derived unit because joule is defined the work done by the force aftab 1 newton causing the displacement of one metre something newton metre(n-m) is also used to measuring work.

6 0
3 years ago
(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

4 0
3 years ago
On which date is the gravitational force between Earth and the moon the greatest?
Genrish500 [490]

The gravitational forces between the Earth and Moon are greatest when the two bodies are closest together. That happens every 27.32 days, when the Moon is at the perigee of its orbit.

Even if this happened at the same time in every orbit, the date would change, because there are not 27.32 days in a month.

But it doesn't happen at the same time in every orbit ... the Moon's perigee precesses around its orbit, on account of the gravitational forces toward the Earth, the Sun, Venus, Mars, and the other planets.

3 0
3 years ago
Help with a length problem
Andru [333]
It would be B

Explanation:
Because if you're not measuring in inches you want to go the next one down other than inches which would be millimeters!(: hope this helps.
6 0
3 years ago
Read 2 more answers
An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?
hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

P = I^2R

make I subject of the formula we have

I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

8 0
3 years ago
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