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babunello [35]
3 years ago
8

Consider two positively charged particles, one of charge q₀ (particle 0) fixed at the origin, and another of charge q₁ (particle

1) fixed on the y-axis at (0,d₁,0). What is the net force \vec{F} on particle 0 due to particle 1?
Express your answer (a vector) using any or all of k, q₀, q₁, d₁, \hat{i}, \hat{j}, and \hat{k}.
Physics
1 answer:
docker41 [41]3 years ago
4 0

Answer:

Explanation: according to Coulomb's inverse-square law is proportional to the square of distance between them and is given by

F=k\frac{q_0q_1}{r^2}

where r is the distance between the charges & k is the Coulomb's constant

k=1/(4*ε_0*π)

k=9*10^9

the distance between the charges in this question is d_1

hence the magnitude of the force exerted by q_0 on q_1 is given by

F=k\frac{q_0q_1}{d_1^2}

due to location of particle 1 above the particle 0 the direction of force is parallel to y axis and in vector form

F=k\frac{q_0q_1}{d_1^2} j

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Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

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Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

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They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

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Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

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It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

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The reference system for her

Body 3

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Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

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   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

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    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

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