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Otrada [13]
7 months 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]7 months 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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If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______
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Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

F = \dfrac{GMm}{r^2}............(1)

new gravitational force

F' = \dfrac{GMm}{r'^2}

now from the given condition

F' = \dfrac{GMm}{(3r)^2}

F' = \dfrac{GMm}{9r^2}

F' = \dfrac{1}{9}\dfrac{GMm}{r^2}

now, from equation (1)

F' = \dfrac{1}{9}F

\dfrac{F'}{F} = \dfrac{1}{9}

now, the change in gravitational force factor is equal to \dfrac{1}{9}

Hence, the correct answer is option A

3 0
3 years ago
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It was Niels Bohr who proposed it
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2 years ago
a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

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1 year ago
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