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miv72 [106K]
3 years ago
12

What is the magnitude of the electric field 36.0 cm directly above an isolated 2.71×10−5 C charge? Express your answer to three

significant figures and include the appropriate units.
Physics
1 answer:
worty [1.4K]3 years ago
3 0

Answer:

36.0 cm directly above an isolated 2.71×10−5 C charge? Express your answer to three significant figures and include the appropriate units.

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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.

3 0
4 years ago
In a simple series circuit, why does the bulb light up when you close the switch
photoshop1234 [79]
<span>closing the switch completes the circuit</span>
8 0
4 years ago
List two examples of energy transformation necessary for the sporting event to take place
Morgarella [4.7K]

Electrical to light & electrical to sound (in commentry) & sometimes battery loudspeakers are used so chemical to sound
5 0
4 years ago
A rock thrown with a horizontal velocity of 20m/s from a cliff that is 125m above level ground. If air resistance is negligible,
Alexxandr [17]

Answer: c

Explanation: 125/20 =6.25

6 0
3 years ago
A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

               u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0
4 years ago
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