An automatic external defibrillator (AED) is simply a series of capacitors used to store a very large charge, which is then discharged through the patient’s chest in a short time. If the capacitor in an AED is fully charged and the AED is no longer connected to the power source, what will happen to the energy stored in the AED if the dielectric (κ = 1.5) is removed?

A) increase by a factor of 1.5

B) increase by a factor of 2.25

C) decrease by a factor of 1.5

D) decrease by a factor of √1.5

**Explanation**

The MCAT will expect you to know a handful of equations for capacitors. The three needed to solve this question are:

C = κε_{o}A / d

C = Q / V

PE = (½)QV

According to the first equation, we see that if the dielectric is removed, capacitance will decrease by a factor of 1.5.

According to the second equation, we see that a decreased capacitance can either be the result of a change in Q or a change in V. In this question, the AED is no longer connected to the battery, so there is no source of additional charges. Thus Q will remain the same and V will increase by a factor of 1.5

Finally, by the third equation we see that if V is increased by a factor of 1.5, the potential energy will increase by a factor of 1.5. Thus (A) is the correct answer.

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