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EastWind [94]
3 years ago
12

You have a system of a positively and a negatively charged objects (+q and -9.) separated by distance d. (a) What is the sign of

the electric potential energy? (b) What can you do to increase the electric potential energy of the system? (C) What can you do to decrease the electric potential energy of the system? Support your answer with a bar chart and mathematically. Specify initial and final states on the bar chart.

Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

a)The sign of energy is negative, b)    we can approach the two charges, c) d ⇒ 0

Explanation:

a) The elective potential energy is

          U = k q₁q₂ / r

Applied to this case.

      U = -k q 9 / d

    U = -9 k q / d

The sign of energy is negative, which corresponds to a stable system

b) c) for the energy to be more negative we can approach the two charges

c) To increase the energetic of the system, that is to say less negative, we can separate the discharges whereby U approaches zero or we can decrease the value of the charge q.

          d ⇒ 0

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Answer:

0.002 N/C

Explanation:

Parameters given:

Charge of object, q = 5 mC = 5 * 10^{-3} C

Acceleration of object, a = 0.005 m/s^2

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> E = \frac{-ma}{q}

Solving for E, we have:

E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C

The magnitude will be:

|E| = |-0.002| N/C = 0.002 N/C

The electric field has a magnitude of 0.002 N/C.

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B. Friction can be a centripetal force, such as when it keeps a car on the road going around a curve.

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The other two options are not correct because:

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