Kaplan FL5 P/C Capacitor Question

[betterfuture]

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A capacitor containing two flat, parallel plates separated by a vacuum is connected to a battery with voltage V. Then, its plates are pulled apart to a distance double the original one. By what factor does the energy density between the plates change?

This is how I worked it out
C=EoA/d, when d doubles, C is halved
U=1/2CV^2, when C is halved, Energy is halved.

Apparently this is wrong. Energy is not halved. The answer they gave was Energy changed by 1/4.(highlite the empty area to show the answer). Please help. Thank you very much.

 

[NextStepTutor_3]

Hi there! On the MCAT, it's vitally important to make sure you answer exactly what the question is asking. Here, the question didn't ask about energy, but energy density. If I remember correctly, this particular passage gave an equation for energy density (which they called "u"), in which it related to electric field strength (E). In this equation, u was proportional to E^2.

Once we see that this exact equation is meant to yield the answer we're looking for, we know we need to be thinking about electric field strength. We can then use V = Ed and the equation they give to find the answer.

Overall, good question! Remember, when the question asks for something you're not 100% familiar with, don't settle for solving for a similar quantity - look back to see if the passage told you how to answer directly.

 

[betterfuture]

Oh sheez. I do see that now. Thank you though. I need to read more carefully!