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3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3

Physics

1 answer:

Pachacha [2.7K]3 years ago

6 0

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC Q₃ = 51 nC Q₂ = 30.5 nC

x₁ = - 1.715m x₃ = - 1.085m x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

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The work done on the car is -20 J.

Work done on the car is negative, meaning that the car actually does work on the external system.

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Energy is the ability to do work
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From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.

Initial energy = 100 J
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Work done on car = 80 - 100J

Work done on car = -20 J

Hence, the work done on the car is -20 J

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2 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

bekas [8.4K]

Answer:

Explanation:

Given that,

Radius r = 15cm = 0.15m

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A = π r²

A = π × 0.15²

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Magnetic field B = 1.2T in positive z direction

B = 1.2 •k T.

If loop is remove from the field in the time interval

∆t = 2.3ms = 2.3×10^-3s

We want to find the average EMF and it is given as

ε = —∆Φ/∆t

The final flux is zero

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Where magnetic flux is given as

Φi = BA Cosθ

Where θ=0 since the area and the magnetic field point in the same direction.

Φi = BA Cos0

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ε = —(Φf — Φi) / ∆t

ε = —(0-0.0848) / (2.3×10^-3)

ε = 0.0848 / (2.3×10^-3)

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3 years ago

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3 years ago

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Answer:

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