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Nuetrik [128]
3 years ago
5

An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.

What are (a) the magnitude of the electric dipole moment and (b) the difference between the potential energies for dipole orientations parallel and antiparallel to Upper E Overscript right-arrow EndScripts?
Physics
1 answer:
bazaltina [42]3 years ago
6 0

Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

                        p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

                      θ = 0º

                      U = 1.16 10⁻¹⁴ 1160 cos 0

                      U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

                       θ = 180º

                      cos 180 = -1

                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

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lakkis [162]

Answer:

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In this exercise on simple harmonic motion we are given the expression for motion

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remember angles are in radians

 

b) The general form of the equation is

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when comparing the two equations

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           f = w / 2π

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      a = - 5,0 2² cos (π / 6)

      a = - 17.31 cm / s²

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3 0
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irakobra [83]

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