# Work Done per Unit Charge vs Current

1. Oct 24, 2013

### san203

A trivial question.
I know that current is the amount of charge passing per unit time across a area.
And Potential difference is change in potential or change in potential energy per unit charge or work done per unit charge.
And also from Ohms law, V is directly proportional to I. So increasing potential difference or voltage should increase current. But how does increasing work done on a charge carrier increase the no. of such charge carriers passing through a given area?

Secondly, any kind of potential actually changes with position,right? So as a electron moves from higher to lower potential, it experiences a potential drop. But why does the potential drop increase across a resistor? After browsing related threads, the answers state things like loss of energy as heat and alike. I agree that the kinetic energy of charge carriers continuously changes due to collision which in turn change into Thermal energy and heat up the resistor, but what relation does this have with a more significant potential drop?(or am i just assuming things here)?

I am hoping to get answers from the Work point of view.

2. Oct 24, 2013

### mikeph

Ohm's law isn't a result of any fundamental dimensional analysis; it's empirical. We happen to notice that the rate of charge passing a cross section of a wire is proportional to the applied potential difference across the wire in certain circumstances, and we call that conductor Ohmic. It's not always the case.

Why does the potential drop increase across a resistor? In the context of this question I think "resistor" should be interpreted as "material with lower conductivity as compared to its connectors". The potential gradient is higher across this material because, as a Coulomb moves from one potential to another, it gains more energy from the electric field when it travels across a material of lower conductivity. This is because more work is required for the electron to be moved through the resistor as compared to a conducting wire.

Think of water flowing in a river. The current in the river is driven by the gravitational potential energy difference between the source (high) and the mouth (sea level). In this analogy, try to think what is the equivalent of a resistor. Let's say the current of the water is constant down the river. Where in a river is the potential drop the highest (per unit path length?). Where in a river does the water gain kinetic energy the fastest? The same place...

In the same way, the charge crossing a resistor gains the most energy and the voltage correspondingly drops much faster.

3. Oct 24, 2013

### Philip Wood

Recall that work is force.displacement. So p.d is force per unit charge.displacement. So potential gradient is equal to electric field strength. So it's not surprising that the free charges move faster if you increase the p.d., because you're increasing the electric field strength. In some circumstances the drift velocity is actually proportional to the applied p.d.

4. Oct 25, 2013

### san203

But again the answer is about moving electrons faster. How does moving electron faster = increase in current, unless of course the no of electrons crossing the area in some time interval increases.

I see.
I have read replies to similar threads. What these replies always stress on is that potential drop only happens at resistors. Don't they also happen as the electron moves away from source?

But before asking any more questions i would like to discuss about the model i have inside my head when i think about circuits and electrons. The model similar to that of a positive test charge present in an electric field created by a positive charge. As the test charge is initially near to the Source, it has high potential energy and hence is at high potential. And because both charges are of same sign, they repel each other. The test charge then moves away from the source to a region of low potential and loses potential energy at every position in its path of motion until it reaches the point of low potential. So similarly in a circuit, the battery produces charges which creates an electric field. A charge carrier then moves from the positive terminal of battery to the negative terminal losing potential energy at every point. So at any two points in the circuit their would be a potential drop. I ran into problems after i tried to think of what would happen to a electron after it hits a resistor. More work would be required to move it out of the resistor and electric field also decreases as it goes farther away from positive terminal.

Unfortunately that was not all. After reading some more, i have come to know that charges at terminals of battery are not the only contributors of electric field. Their are charges along the surface which are a major source of electric fields. And that at the interface of a conducting wire and resistor, their are electrons which produce a strong electric field within the resistor .
This brings up some questions

1.) As their are multiple electric fields, will electric field strength decrease as the charge carriers moves farther away from source like it used to?

2.) And how will these additional electrostatic fields affect electric potential

5. Oct 25, 2013

### mikeph

I think you're over-thinking this. The potential drop across a resistor is greater because more energy is needed to move a coulomb through a resistor than through a conducting wire.

6. Oct 25, 2013

### Philip Wood

That's just it: it's easy to show that if free electrons are the only carrier, then if there are $\nu$ of them per unit volume in a conductor of cross-sectional area A, then $I = \nu A v e$.

Turning to resistance, it's useful to define the free electron mobility, $\mu$ by the equation $v = \mu E = \mu \frac{\Delta V}{\Delta x}$, where this scalar version is fine for an isotropic conductor. Velocity rather than acceleration is involved, because after E is applied the electrons quickly reach a mean drift velocity, v, owing to collisions with the vibrating ions. So we see that $R = \frac{\Delta V}{I} = \frac{\Delta x}{\nu \mu eA}$. I've gone through this (albeit hurriedly) because it sets the scene for how resistance is thought of on a microscopic scale. It dodges the difficult issue of how $\mu$ can be calculated theoretically. Note that if $\mu$ is a constant, independent of E, then the conductor is ohmic.

Another issue you've raised which I can't answer properly is the origin of the electric field in a resistor. It's clearly set up by unbalanced charges (say on the terminals of a battery), but it doesn't fall off with an inverse square law as you go through a resistor connected across the terminals. Instead, for a uniform resistor the field is uniform as we go through the resistor, because the current is constant all the way through. I don't find this easy to picture in terms of charge distribution. But maybe this is my problem, not yours.

Last edited: Oct 25, 2013
7. Oct 28, 2013

### Philip Wood

San203: Are you any further forward?

8. Oct 29, 2013

### san203

@Phillip Wood i didnt get you? are you asking about my findings to the question?

I have another question. Consider a simple circuit with a battery and a resistance. i have - as mentioned in my previous post - always thought that the potential of a charge carrier drops as it goes across the circuit. But i saw a Potential vs Position graph in my textbook and their was absolutely no change in the slope from end of resistor to the zero potential terminal of the battery? Does this mean that after a voltage drop across the resistor, no potential difference in the wire exists?

9. Oct 29, 2013

### Philip Wood

Sorry to cause confusion. I simply meant to ask whether my reply (6), or any other enquiries you made, or thinking you've been doing, have led to progress in finding an answer.

The standard answer to your question is to appeal to V = IR, and the fact that current is the same throughout the circuit (otherwise charge would pile up etc). For our purposes V = IR is better written as ΔV = I ΔR, that is the potential drop, ΔV, across an element of circuit with resistance ΔR is proportional to ΔR, since I is constant. That's why there's a steady decrease in V as we go through a resistor, but a negligible ΔV as we go through a connecting wire, as its resistance is negligible.

If I've understood you aright, you are looking for something more than this, for an answer in terms of the way the charges distribute themselves in the wire, in order to give these potential gradients - constant in the resistor, negligible in the connecting wire. As I've said earlier, I think this is an interesting and valid question, but not one which I've ever seen discussed. All you usually get is the answer I gave in the last paragraph. Let's hope someone brainier than I takes up your challenge!

10. Oct 29, 2013

### san203

1.) I seem to have found out the answer to my problems after reading a few books.
Summary -
I have learned that in an electrical circuit, the battery isn't the only source of charges. The electric field by the charges at the terminals first creates an intermediate electric field which in turn moves other electrons in the wires to pile up at bends(surface charge) so that electric field points in direction of the wire.
As for the resistor, during the initial current, charges pile up at resistor. Thats because the amount of charge entering it less than amount of charge leaving it. This process continues until the electric field inside the resistor is strong enough for constant current through out the resistor. And because we have stronger electric fields in resistors, the potential drop across the resistor is higher.

This was what i was looking. I don't know if this helped you although i was the one asking this question.. lol.

2.) The answer to the question i asked in the previous post is as you said. Low resistance, low eletric field, low change in potential but definitely not zero.

I got the answers from Matter and Interaction by Chabay and Sherwood. Good book if people want microscopic understanding of Circuits but it has only one chapter on it.

11. Oct 29, 2013

### Philip Wood

Thanks for the reference. In my last post I should have said "otherwise charges would continue to pile up". I was referring to the steady state, rather than the initial establishment of steady flow.