Question

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same? C decrease Correct: Your answer is correct. Q decrease Incorrect: Your answer is incorrect. E between the plates stay the same Correct: Your answer is correct. ΔV increase Correct: Your answer is correct.

Answer 1

Q: remains the same

C: decreases

ΔV: increases

Step-by-step explanation:

When the capacitor is disconnected from the battery, and the wire connected to the plates are not touching anything else, it means that the charge cannot flow out from the capacitor: so, the charge stored on the plates of the capacitor, Q, will not change, regardless of the distance between the plates.

The capacitance of a parallel plate is given by

`C=(\epsilon_0 A)/(d)`

where A is the area of the plates and d the separation between the plates. As we see from the formula, C is inversely proportional to d: so, if the plates are pulled apart to a larger separation, it means that d increases, and so C decreases.

Finally, the voltage across the capacitor is given by

`\Delta V=(Q)/(C)`

and since we said that Q does not change while C decreases, it means that
`\Delta V` increases, since
`\Delta V` is inversely proportional to C.