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Otrada [13]
1 year ago
11

At what distance from the wire is the magnitude of the electric field equal to 2.41 n/cn/c ?

Physics
1 answer:
Sladkaya [172]1 year ago
5 0

The correct answer is 1.07m.

The area surrounding an electric charge where its impact may be felt is known as the electric field. When another charge enters the field, the presence of an electric field may be felt. The electric field will either attract or repel the charge depending on its makeup. Any electric charge has a property known as the electric field. The charge and electrical force working in the field determine the strength or intensity of the electric field.

Here, is the charge per unit length, r is the distance from the wire, and

is the free space permittivity  ε_0. Electric field due to the long straight wire is,

E= λ/2πε_0r

Rearrange the equation for r.

r=λ/2πε_0E

Substitute 2.41 N/C for E,

E=1.44×10^-10C/m

λ=8.85×10^-12C^2/Nm^2

r=(1.44×10^-10C/m)/(2(3.14)(8.85×10^-12C^2/Nm^2)(2.41N/C))

r=1.07m

At a distance of 1.07 m the magnitude of electric field is 2.41 N/C.

To learn more about electric field refer the link:

brainly.com/question/12821750

#SPJ4

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denis23 [38]

Answer:

4.245s

Explanation:

Given that,

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t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

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Therefore, the time the driver measures for the trip is 4.245s

8 0
3 years ago
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zmey [24]

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Science whoever gets this will get a brainlest
Zanzabum

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I got it right xd

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