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2 years ago

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A charge of +Q is fixed in space. A second charge of +q was first placed at a distance r1 away from +Q. Then it was moved along

a straight line to a new position at a distance R away from its starting position. The final location of +q is at a distance r2 from +Q.
What is the change in the potential energy of charge +q during this process?
(A) kQq/R
(B) kQqR/r12
(C) kQqR/r22
(D) kQq((1/r2)-(1/r1))
(E) kQq((1/r1)-(1/r2))

1 answer:

olga55 [171]2 years ago

5 0

Answer:

Option (D)

Explanation:

The formula for the potential energy between the two charges is given by

$U=\frac{KQq}{r}$

where, r is the distance between the two charges.

In first case the distance between the two charges is r1.

The potential energy is

$U_{1}=\frac{KQq}{r_{1}}$

In first case the distance between the two charges is r2.

The potential energy is

$U_{2}=\frac{KQq}{r_{2}}$

The change in potential energy is

$\Delta U = U_{2}-U_{1}$

$\Delta U=\frac{KQq}{r_{2}}-\frac{KQq}{r_{1}$

$\Delta U=KQq \times \left ( \frac{1}{r_{2}} -\frac{1}{r_{2}} \right )$

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Explanation:

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A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure

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Answer:

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Explanation:

Hello!

To solve this problem we must use the following steps

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α=

metal rod resistivity=2x10^-4 Ωm

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solving

$R=(2x10^-4)\frac{2}{0.00283} =0.14$

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

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I1+I2=5

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solving for i1

I1=14285.7i2

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14285.7i2+i2=5

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2 years ago

A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction o

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2 years ago

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Answer:

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2 years ago