# Why is electric field larger in a resistor?

Suppose there is wire with 0 resistance and a resistor between them and that circuit is plugged to nonzero voltage source. Why is the electric field in resistor larger than of the wire? I want to think in terms of force. When electron enters from wire to resistor I think electron will slow down as a result this affects electrons behind them so the force will be decreased from top to bottom which I think will imply electric field will be uniform across the wire and resistor.

For source of the result I mention in question:

At 0:15 in

the instructor explains that in resistor there will be larger acceleration for electron between relaxing time implying there is larger electric field around that region. As instructor says that drift velocity is constant as time passes then in resistor there will be larger acceleration for electron but I think drift velocity is overall velocity of electron not just a single electron which to me doesn't makes sense.

This can be shown from the combination of $$V=IR$$ and $$E=V/l$$ where $$V$$ is the voltage, $$I$$ is the current, $$R$$ is the resistance, $$E$$ is the magnitude of the E field, and $$l$$ is the length of the wire or resistor.
By substitution we immediately get $$E=IR/l$$. We always have $$l>0$$, and for series connected components the $$I$$ in each component is equal to the $$I$$ in the other component. So for $$R=0$$ we get $$E=0$$ and for $$R>0$$ we get $$E>0$$. Thus the E field in a resistor is larger than the E field in an ideal conductor.
Force is not necessary for this. You could introduce it, since $$E=F/Q$$, but it doesn’t simplify anything. You simply tack the force on to the end after you have already shown the difference in the E field. Similarly with drift velocity and relaxation time. They don’t help clarify the E field, but knowing the E field then you can easily get the rest.