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My name is Ann [436]
3 years ago
13

Please help!!

Physics
2 answers:
Luba_88 [7]3 years ago
8 0

Answer:

Bath CD jshchdhdhfhfhhfhd jpg de f for frr for gi Jhong GO by be jr jpg be

erastova [34]3 years ago
7 0

34 bc of the velocity of the arrow will come down with less dense

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A long, thin straight wire with linear charge density ? runs down the center of a thin, hollow metal cylinder of radius R. The c
MaRussiya [10]

Answer:

a. E = \frac{\lambda }{(2\epsilon * r) }

b. E = \frac{3(lambda)}{2\pi*\epsilon(outside)  }

Explanation:

The important thing to remember is to use Gauss Law. This is a relation that describes the distribution of electric charge to the resultant electric field.

Linear charge density means charge per unit length of material.

Data:

The metal cylinder is hollow.

The unit length is L.

a.The expression will be as follows:

for charge inside the cylinder, where r < R, the expression is:

E = \frac{\lambda }{(2\epsilon * r) }

b. Let's assume that the cylinder is a coaxial cylinder with a radius r > R, then the electrical field strength is given as:

E = \frac{Q ( enclosed)}{A*\epsilon }

E = \frac{\lambda*L+2(\lambda)*L }{(A)*\epsilon(outside) }

E  = \frac{3(\lambda)L }{(2\pi*R*L*\epsilon(outside)  }

This gives:

E = \frac{3(lambda)}{2\pi*\epsilon(outside)  }

The solution informs us that there is a surface change taking place on the cylinder. Therefore, there will not be a magnetic field across it.

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