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If the electric circuit is complete, then one can say that the net average displacement of the electrons is $0$. Then consequently the work done by the electric field is also $0$.

So how is energy supplied to an electrical appliance for example since the electric force is doing $0$ work?

I think this has something to do with the Poynting vector or the fact that I just can't say that the displacement of the electron is $0$...

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Consider the left hand circuit.

enter image description here

Evaluating the work done by the electric fields in taking unit positive charge from node $A$ and moving clockwise around the loop and back to $A$.

$$\displaystyle \int^B_A \vec E_1\cdot d \vec l + \int^A_B\vec E_2\cdot d \vec l=0$$

This is Kirchhoff's voltage law in going around a complete loop.

Interpreted another way this equation can be written as

$$\displaystyle \int^B_A\vec E_1\cdot d \vec l = - \int^A_B\vec E_2\cdot d \vec l = + \int^B_A\vec E_2\cdot d \vec l$$

All this equation is saying is that $v$, the potential difference between $A$ and $B$, is the same whether one goes via resistor $R_2$ in which case $v = - \displaystyle \int^B_A\vec E_1\cdot d \vec l$ or one goes via resistor $R_1$ in which case $v = - \displaystyle \int^B_A\vec E_2\cdot d \vec l$.

There is some external source which is driving a total current $i=i_1+i_2$ through the resistors and doing work to drive the currents through the resistors wth the result that there is a heating effect within the resistors.


To consider where that work might come from look at the right hand diagram and starting at $A$ and moving in a clockwise direction consider the work done by the electric fields on a unit positive charge.

$$\displaystyle \int^Y_X \left (-\vec E_{\rm C}\right )\cdot d \vec l + \int^X_Y\vec E_{\rm R}\cdot d \vec l=0$$

The first term is negative and so the work done by the electric field is negative which is the same as saying that some external force acting on the unit positive charge has done work.
This work has been done as a result of the chemical processes happening within the cell and this is where external work is done on the electrical circuit.
The second term is positive and here the work done by the electric field in driving unit positive charge through the resistor produces a heating effect within the resistor.

Overall the energy changes might be thought os as follows: $\rm chemical \rightarrow electrical \rightarrow heat$

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  • $\begingroup$ I did not understand your math,how did you write integral from x to y (-Ec.dl)?According to me it should be integral from y to x (-Ec.dl) $\endgroup$ – Schwarz Kugelblitz Sep 22 at 11:11
  • $\begingroup$ @SchwarzKugelblitz The path taken is from $X$ to $Y$ through the cell and then from $Y$ back to $X$ through the resistor. $\endgroup$ – Farcher Sep 22 at 11:20
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The problem is in the word “average”.

If you push a heavy box back and forth, the average displacement of the box is zero, but the sweat on your body shows that your total work is not zero, but that it is the sum of works.

Work is summed as scalars, not as vectors.

Another problem may be with confusing the total work with the law of conservation of energy. Total energy will be the same, but one body may lose it (a battery) because of doing work (heating a resistor, rotating a motor, etc.), while other body may receive it (heat, kinetic energy, etc.).

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  • $\begingroup$ But if I push a box back and forth and bring it back to the place I started from the work would be 0 because the displacement is 0 $\endgroup$ – Schwarz Kugelblitz Aug 11 at 15:56
  • $\begingroup$ Did you try it? Do you have the feeling that your work was zero after it? It seems for me that you mix the law of conservation of energy with the work. Total energy will be the same after pushing the box back, but you performed a work, and so you lost energy, which was transformed to the heat (from the friction). $\endgroup$ – MarianD Aug 11 at 16:08
  • $\begingroup$ Hold on...the definition of work is F.S where s is the displacement which is 0 in my case... $\endgroup$ – Schwarz Kugelblitz Aug 11 at 16:10
  • $\begingroup$ U are saying that the total energy remains constant after bringing the box back to the original position but later said that I lost energy to friction? $\endgroup$ – Schwarz Kugelblitz Aug 11 at 16:11
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    $\begingroup$ @Aaron, from the biological point of view, a shell with the smooth muscle does no work, but me, a human being with the skeletal muscle, do a work, as my individual muscle threads permanently stretch and release. The heavy box does acquire (almost) zero energy from me, but my chemical energy transfers to the thermal one. $\endgroup$ – MarianD Aug 11 at 17:37

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