# AAMC qpack physics #24

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#### poochimaster

##### Full Member
7+ Year Member

So the answer is A, and the explanation is shown here. Now I get why increasing L by a factor of 2 will decrease the electric field (via formula E = V/L), but I don't get why decreasing R by a factor of 2 does not have the same effect.

My rationale is: via V = IR, if you hold current constant decreasing R by a factor of 2 would decrease V by a factor of 2. If V decreases by a factor of 2, E would also decrease by a factor of 2.

So, where did the AAMC solution get the formula E = (V - IR)/L from? What error am I making in my reasoning?

Thanks!

What circuit?

The battery holds the voltage constant. Decreasing R would increase I, not hold it constant.

What circuit?

Sorry for being vague but the question doesn't refer to a circuit in specific, just circuits in general.

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The battery holds the voltage constant. Decreasing R would increase I, not hold it constant.

Ahh gotcha! Makes sense. Can you explain to me where AAMC got that formula from though, the E = (V - IR)/L?

Thanks!

The circuit setup has a battery, resistor, and capacitor (vacuum photodiode). The voltage from the battery drops across the circuit elements in one closed loop, with these elements being the resistor and capacitor.

The voltage drop across the resistor = IR. The remaining voltage drops across the capacitor which is V - IR. (V is the voltage from the battery).

The relationship between E-field and voltage between plates of a capacitor is V = E * L where L is the distance between the plates (anode and cathode). So in this case E = (V - IR) / L.

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