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I have read that the analogy of gravitational field is similar to electric potential energy..now if I have a ball falling from higher potential level to lower potential level,then the potential energy of ball changes to kinetic energy. Similarly will the potential energy of a charge gradually decrease and change to kinetic energy as it moves through a conductor under application of potential difference My teacher told that the potential energy of charge remains constant until it gets obstructed by a resistor.. I am confused that this contradicts the analogy of gravitational potential energy as I discussed earlier

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  • $\begingroup$ What's the sign of the charge? $\endgroup$
    – Qmechanic
    Commented May 29, 2020 at 15:59
  • $\begingroup$ It's positive (conventional) $\endgroup$ Commented May 30, 2020 at 4:13

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Does potential energy of a charge gradually decrease when it moves from a higher potential level to low potential level?

Yes. Suppose you have an electron in the space between the anode and cathode plates of a vacuum tube. The charges on the two plates will set up an electric field in the space between them, and there will be a potential gradient associated with that field.

As the field accellerates the electron, it will move from the lower potential area to the higher potential area, losing electrical potential energy (because of its negative charge), and gaining kinetic energy.

My teacher told that the potential energy of charge remains constant until it gets obstructed by a resistor.

Now you're talking about a different situation.

Suppose you have a battery (electrochemical cell) with a resistor attached across its terminals by two wires.

Now the wires have very low resistance, and their presence will rearrange the electric fields so there is very little electric field within the wires, and so very little potential difference between one end of a wire and the other.

As charge moves along the wire, it does lose potential energy. It doesn't tend to gain significant kinetic energy because it interacts with the atoms that make up the wire, transferring any gained kinetic energy to them and producing heat in the wire. The wires are themselves resistors, just with much lower resistance than the circuit element we normally call a resistor.

However, because the resistance of the wire is much smaller than that of the resistor, and so the potential drop along the wire is often very much smaller than the potential drop due to the resistor, we often make a useful simplifying approximation and say that the potential drop along the wire is zero. This can greatly simplify solving many circuits, without introducing a large enough error significantly alter our results.

Of course we should be mindful of the limits of this approximation, and beware that sometimes the wire resistance might be significant, and we'll need to consider it to solve certain circuits accurately enough (for whatever purpose we're solving them).

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  • $\begingroup$ Love the way you explained it..but can you elaborate the idea of saying ''Now the wires have very low resistance, and their presence will rearrange the electric fields so there is very little electric field within the wires, and so very little potential difference between one end of a wire and the other.'' $\endgroup$ Commented May 30, 2020 at 4:05
  • $\begingroup$ @Curious.SWARNIM, you should read the dozens of previous questions on this topic that have been posted and answered here. $\endgroup$
    – The Photon
    Commented May 30, 2020 at 4:27
  • $\begingroup$ Okk I will check it out $\endgroup$ Commented May 30, 2020 at 4:33
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Consider a simple circuit: an ideal DC battery with emf $V$, a resistor with resistance $R$ and ideal connecting wires.

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Let us assume that the potential at the positive terminal to be $V$ volts and that at the negative terminal as $0$ volts. Now, consider the end of the connecting wires near the resistor (at point b). The potential at point b is also $V$, since the connecting wires are ideal (they have a resistance of $0$ $ \Omega $, and the current flowing is constant $i=\frac VR$, hence, in accordance with Ohm's law, if $R$ is zero, then, so is the potential difference). So, as you can see, there is no potential difference here.

This is similar to rolling a ball on a plane ground. The ball keeps rolling (without any loss in its potential energy), and at the resistor it reaches a downward slope, where its potential energy is lost. For the next cycle, the body has to be supplied the potential energy again, which the battery does.

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Similarly will the potential energy of a charge gradually decrease and change to kinetic energy as it moves through a conductor under application of potential difference.

Yes, under just the influence of an electric field charges will move in such a way that their electric potential energy decreases. For positive charges this means moving from higher to lower potential, and for negative charges this means moving from lower to higher potential.

My teacher told that the potential energy of charge remains constant until it gets obstructed by a resistor.

The context here is different than the previous discussion. Your teacher is most likely talking about charges in an electric circuit. Here, to good approximation, we assume that wires in a circuit are perfect conductors with no resistance. Therefore, there is no electric field in the wires, and so there is no potential change as you move along the wire. Therefore, the potential energy of a charge moving along an ideal wire will not change.

This is different than the previous scenario where we assumed the charge was moving in the presence of an electric field.

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I have read that the analogy of gravitational field is similar to electric potential energy..now if I have a ball falling from higher potential level to lower potential level,then the potential energy of ball changes to kinetic energy.

It depends on the specifics of the gravity analogy.

If the ball is in free fall (no air resistance) in a gravitational that would be equivalent to a charge in a vacuum (encountering zero resistance) under the influence of an electric field which accelerates the charge increasing its kinetic energy at the expense of electrical potential energy. An example is the old cathode ray vacuum tube in TVs where electrons are accelerated by an electric field.

But the analogy does not apply when electrical resistance is involved. The charge continuously collides with the atoms and molecules of the resistance alternatively gaining kinetic energy from the field and losing it in the collisions, resulting in resistance heating. The charge moves at a constant average drift velocity (constant current) through the resistance. In this case, the gravity equivalent is an object falling in air that reaches its terminal velocity due to air resistance when the upward force of the air resistance eventually equals the downward force of gravity for a net force of zero. From there on, as the object falls at constant terminal velocity its loss of gravitational potential energy is converted to air resistance heating. There is no increase in the velocity (kinetic energy) of the object.

Similarly will the potential energy of a charge gradually decrease and change to kinetic energy as it moves through a conductor under application of potential difference

As already indicated above if the conductor has resistance there will not be a net increase in kinetic energy of the charge. The electrical potential energy is converted to resistance heating.

My teacher told that the potential energy of charge remains constant until it gets obstructed by a resistor..

Your teacher is talking about an ideal (zero resistance) conductor which generally don't exist (except for super cooled conductors). All conductors have some resistance. But we normally regard the resistance of wires that connect together circuit components as being so small compared to the resistance of the circuit components themselves, that the resistance of the conductors can be ignored (considered zero).

Hope this helps.

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  • $\begingroup$ Well, that's good for potential energy of charge but can you tell me how does potential changes throughout the circuit (I know that potential is independent of charge) $\endgroup$ Commented May 30, 2020 at 4:18
  • $\begingroup$ @Curious.SWARNIM Well, it was potential energy you asked about, wasn't it? Anyway, for conventional current, which is defined as the flow of positive charge, and the direction of the electric field, which is defined as the direction of the force that a positive charge would experience in the field, then (1) for positive charge an increase in potential energy is also an increase in electrical potential and (2) for negative charge (which flows in opposite direction) an increase in potential energy is a decrease in electrical potential. $\endgroup$
    – Bob D
    Commented May 30, 2020 at 14:01
  • $\begingroup$ You mean that as the positive charge moves from positive terminal to negative terminal the potential at each point along with potential energy also decreases(ie. Potential decreases as it flows towards negative terminal and a point comes when it becomes zero , and as it continues potential becomes negative till it reaches the terminal)...right? $\endgroup$ Commented May 30, 2020 at 15:11
  • $\begingroup$ @Curious.SWARNIM Yes, at each point going through the resistor from positive potential on one side of the resistor to more negative potential, the potential and potential energy decreases. But it doesn't necessarily go to zero on the other side of the resistor. It depends on where zero potential is assigned in the circuit. In the case of a simple circuit consisting of a single resistor connected across a battery, the negative side is at zero potential. If there are two resistors in series with the battery, the resistor connected to the negative battery terminal would be at zero potential. $\endgroup$
    – Bob D
    Commented May 30, 2020 at 15:58

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