AAMC physics question pack 24

[bellowbruins]

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So, the question asks which of the following changes to the circuit will decrease the electric field between teh electrodes by the greatest amount?

A. Increasing L by a factor of 2

B. Decreasing L by a factor of 2

C. Increasing R by a factor of 2

D. Decreasing R by a factor of 2.


I got the answer correct but when i read the solution i dont know where does E= (V-IR)/ L comes from. I only know E= V/L or at least i can substitute ohm's law E= IR/L. I dont see that V-IR coming from. lol please help.

 

[desperadoflakes]

The electric field for this problem is equal to the delta V divided by L.

So E = (V final - V initial)/L
They decided to call V initial as being = IR I guess? Ohms law. Do you have a screenshot of the problem?

 

[rabbott1971]

Here is the problem. I also got it right but not super sure how.

 

[desperadoflakes]

I think we're supposed to think about the electrodes as being akin to the parallel plates of a capacitor. Even if you increase the resistance and decrease the current, eventually you're still going to have the same build up of charge regardless. So really the electric field is just a function of the capacitance, and we know that increasing the distance between plates decreases the capacitance???

 

[rabbott1971]

I can see from their equation how it works mathematically, but I didn't know the equation, and probably won't at test time. But I think you're right about capacitance, since it is determined by geometric factors (plate size, L).

 

[desperadoflakes]

Yeah, they're definitely talking about it as analogous to a capacitor. The equation that they give is the equation you use to find the voltage across a capacitor in a series where there's also a resistor.

If there was no resistor there, then the voltage across the plates would be equal to the voltage of the battery (EMF). But since there is a capacitor, the voltage is equal to the voltage of the EMF - the voltage across the resistor. According to Ohm's Law, the voltage in the resistor is IR.

So voltage in a capacitor that is in series with a circuit is = EMF - IR.

Because E = V * distance between plates L, E = (EMF-IR)/L