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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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5 0
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7 0
3 years ago
Read 2 more answers
A frog is at the bottom of a 17-foot well. Each time the frog leaps, it moves up 3 feet. If the frog has not reached the top of
docker41 [41]

Answer:

The frog takes 8 jumps to reach top of well

Explanation:

Given data

Frog at bottom=17 foot

Each time frog leaps 3 feet

Frog has not reached the top of the well, then the frog slides back 1 foot

To Find

Total number of leaps the frog needed to escape from well

Solution

in 1 jump distance jumped=3+(-1)

                                           =2 feet

                                           =2×1 feet

The "-1" is because the frog goes back

Now After 2 jumps the distance jumped as:

                     Distance Jumped=2+2

                     Distance Jumped=2*2

                                                   =4 feet

Similarly after 7 jumps

                    Distance Jumped=2+2+......+2

                    Distance Jumped=2*7

                                                 =14 feet

Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.

So

              Distance Jumped=(Distance Jumped after 7 jumps)+3

                                           =14+3

                                           =17 feet

The frog takes 8 jumps to reach top of well                

7 0
2 years ago
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