Physics Question

[PanRoasted]

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The problem is "The capacitor is an automatic external defibrillator delivers 100 J of energy at a voltage of 1000V. How much energy would be delivered at 1500V?"

Supposedly, 1/2 C V^2 = 1/2 Q V = W, but for some reason when I try to solve this using the relationship 100J=1/2 Q V to solve for Q, and then use that Q to solve for work at 1500V, I get a different answer than when I solve for capacitance and then use that to find the work at 1500 V. What huge concept am I missing here? Thank you.

For the record, using capacitance gives an answer of 225 J while using the other formula gives an answer of 100 J.

 

[SN2ed]

Thread moved. These types of questions belong in the MCAT Study Q&A forum.

 

[ilovemcat]

Think about it this way: If you increased the voltage on a parallel-plate, would the charge on it increase, stay the same, or decrease?

Assuming you're using the same plates, the capacitance itself wouldn't change. It's value is dependent on the area of the plates (which it's proportional to), and the distance between them (inversely proportional). Therefore, if capacitance (C) isn't changing, but you increase the voltage (V), charge in turn would have to increase as well, since they're proportional (Q = CV). In your calculation, you made the mistake of assuming charge would remain the same (by increasing voltage), which isn't true at all.

The easiest way to solve this problem is to ask yourself, "What isn't changing in both scenarios?" And the answer to that is capacitance. That's what you need to solve for. Once you solve for capacitance, plug that into the equation: 1/2CV^2. This equation will give you the potential energy stores by the plates. This energy will eventually be used to power the defibrillator.

 

[GreenRabbit]

The problem is "The capacitor is an automatic external defibrillator delivers 100 J of energy at a voltage of 1000V. How much energy would be delivered at 1500V?"

Supposedly, 1/2 C V^2 = 1/2 Q V = W, but for some reason when I try to solve this using the relationship 100J=1/2 Q V to solve for Q, and then use that Q to solve for work at 1500V, I get a different answer than when I solve for capacitance and then use that to find the work at 1500 V. What huge concept am I missing here? Thank you.

For the record, using capacitance gives an answer of 225 J while using the other formula gives an answer of 100 J.

By using the first equation you're holding capacitance constant while letting charge change with voltage. By using the second equation you're holding charge constant while letting capacitance change with voltage.

Since capacitance is intrinsic to the physical design of the capacitor, you should use the first equation.