# AAMC physics question pack 24

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#### bellowbruins

##### Full Member
5+ Year Member
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.

##### Full Member
10+ Year Member
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

##### Full Member
7+ Year Member
Here is the problem. I also got it right but not super sure how.

#### Attachments

• AAMC Physics QPack #24.JPG
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##### Full Member
10+ Year Member
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???

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#### rabbott1971

##### Full Member
7+ Year Member
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).

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