AAMC Exam4 P.S.#41

[Labminion]

For those of you who have taken this exam, I have a question about #41 on the physical science section of AAMC practice exam #4. If you don't have access to the exam, the passage was about the photoelectric effect, and the figure showed two electrodes (basically, a capacitor) in a vacuum connected to a voltage source. Connected in series with the electrodes is a resistor. When a photon hits the cathode of the capacitor, if it has sufficient energy, an electron is liberated and accelerates through the electric field created by the capacitor.

Ok. The question is: which of the following changes to the circuit will decrease the electric field between the electrodes by the greatest amount? The correct answer (which I agree with) was: increasing the distance between the plates by a factor of 2. One of the other answers involved increasing the circuit resistance by two. I figured, since no current is moving through the circuit while the capacitor is charged, that increasing resistance by 2 would have no effect on the electric field. However, the answer key claims increasing the resistance will decrease the electric field, just not to the same extent that increasing the plate distance by two would. Isn't the voltage across the charged capacitor equal to the battery voltage? Or do we have to subtract the voltage drop across the resistor, even though no current is flowing?

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[liveoak]

internal resistance perhaps?

if you increases resistance, you decrease current, and thus you increase the terminal voltage. which would increase the electric field....never mind.

 

[geeyouknit]

I always thought that voltage drops across a resistor, ie, if you just had a circuit with a battery and a resistor, the battery being 5 volts, the resistor has to decrease the voltage (not sure if that's correct terminology) by 5 volts for Loop rule to be correct.

If there's a capacitor, it would still be the same thing. If current has to flow across the resistor, then the resistor will cause the voltage to drop by a certain amount. The amount of voltage that remains is what charges the capacitor. Even for the capacitor to fully charge, the current has to cross the resistor to get to the cap.

Can someone else verify this/answer OP's question? My answer makes sense to me, but I have a great talent in rationalizing things the wrong way.

 

[paul411]

The AAMC explanation is fairly clear

You first have to know that the electric field across a capacitor is: E = V/d

Now, if we add a resistor to this circuit, the voltage drops across the resistor, so the voltage across the capacitor will also decrease. By how much will it decrease? current (I) times the resistance (R) = IR. So the new voltage is V-IR. So across the capacitor, we have E = (V-IR)/d (which is what the AAMC explanation gives).

Now that you have the formula, you can see how increasing resistance or d will impact the electric field.
Increasing R by 2 gives E = (V-2IR)/d = V/d - 2IR/d
Increasing d by 2 gives E = (V-IR)/2d = V/2d - IR/2d

Looking at the equations, it should be clear that given V and I are constant, increasing d by a factor 2 will decrease E more than would increasing R by a factor of 2.

 

[Labminion]

Thanks for the input all. I did review AAMC's explanation, but to me it is still not clear why the resistor would decrease the voltage by a factor of V=IR if there is no current flowing (the capacitor is fully charged, so it is like a broken circuit). No current, I=0, so V=0 across the resistor. Right?

With regards to the comment that I is constant, it isn't though, if I understand the process correctly. Current decreases through the circuit while the capacitor is being charged, and is equal to zero when the capacitor is fully charged. After that, it is dependent on the number of photons hitting the cathode of the capacitor, which is proportional to the number of electrons liberated from that plate. Please correct me if I am wrong. Circuits have never been my strong suit.

 

[paul411]

Ahh, I see the confusion. You reasoning is correct but what you have to understand/infer is that the question is taking about doubling the d or R at any given instant.

 

[Labminion]

Ahh, I see the confusion. You reasoning is correct but what you have to understand/infer is that the question is taking about doubling the d or R at any given instant.

Thanks, you're right. The question did state the voltage across the capacitor was "close" to the battery voltage, which does not mean "equal to". I shouldn't have assumed no current was flowing.