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Sergio [31]
3 years ago
6

Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen

ter of the spherical shell, then the electric field between the two, as well as the field outside the outer boundary of the conductor, is the same as the field you would measure if the conducting shell was not there, though the charges of the conductor will redistribute themselves to ensure zero E field inside the conductor.
True or False?
Physics
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer: True

Explanation: A neutral spherical conducting shell, has no net electric field inside it. The neutral conductor separates its positive and negative charges, when it is kept in a region of electric field, so that the net electric field inside the conductor becomes zero

Let us assume that a spherical Gaussian sphere surrounding the cavity and inside the conductor.

Since electric field inside the conductor doesn't exists, therefore the net electric flux through the Gaussian surface is zero.

From Gauss's law, when net electric flux through the closed surface is zero, the net enclosed charge should be zero

In order to make net enclosed charge as zero inside the metal, the interior surface of the conductor acquires a charge of -q

Since the interior surface of the conductor acquired -q charge, in order to maintain the electrical neutrality of the conductor, the exterior surface of the conductor acquires +q charge on it

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You wad up a piece of paper and throw it into the wastebasket. How far will
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stripped down projectile situation.

You said "an angle of 36 degrees", but you didn't say relative

to what.  I'll assume that it's 36 degrees above horizontal, and

now I'll proceed to answer the question with the information that

I just gave myself.

-- The vertical component of the velocity is  1.4 sin(36)

                                                                        = 0.823 m/s up.

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                                  2 (0.823/9.8) = 0.168 second

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-- The horizontal component of the velocity is  1.4 cos(36)

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-- During the 0.168 second that it's in the air,

the wad travels horizontally

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                                          =            0.19 meter

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If you find my mistake on this one, please please tell me.  

As of now, it looks like with that velocity at that angle, your

paper wad only makes it 7.5 inches from your hand into the can.

Explanation:

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