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Luden [163]
3 years ago
12

Consider an unknown charge that is released from rest at a particular location in an electric field so that it has some initial

electric potential energy. In what direction will the charge move in regards to its potential energy?
A) a positive charge will move to regions of lower potential energy, but a negative charge will move to regions of higher potential energy
B) a charge will move to regions of lower potential energy, regardless of whether it is positive or negative.
C) a negative charge will move to regions of lower potential energy, but a positive charge will move to regions of higher potential energy
D) a charge will move to regions of higher potential energy, regardless of whether it is positive or negative.
E) both positive and negative charges can move toward either higher or lower potential energy
Physics
1 answer:
sineoko [7]3 years ago
4 0

Answer:

b)

Explanation:

If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.

For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)

For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)

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Answer:

Explanation:

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In this way it is possible to say that the energy of a signal is closely related to its amplitude, but its development over time is also important.

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3 years ago
⦁ A certain resistor is required to dissipate 0.25 W, what standard rating should be used?
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Explanation:

6 0
3 years ago
If Vector A is (3, 0) and Vector B is (-3, 3), what is the resultant?
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6 0
3 years ago
What do biologist geologist have in common how are they different
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2 years ago
An airplane accelerates from rest down a runway at 3.20 m/s2 for 32.8 seconds until it lifts off the ground. Determine the dista
vagabundo [1.1K]

Answer:

s=1721.344m  ,v=104.96m/s.

Explanation:

using thr equation of motion;

s=ut+\frac{1}{2}at^{2}

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s=\frac{1}{2}at^{2}

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