Multiple Choice
What is the smallest possible capacitance of a parallel-plate capacitor with plate area separated by a distance , if you can fill it with any of the materials listed in Table 21.3?
26. Capacitors & Dielectrics
Parallel Plate Capacitors
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A parallel plate capacitor is charged up using a battery. The battery is then removed, and the plate separation is increased by somebody wearing insulating gloves. What happens to the potential difference between the plates?
A
It will decrease.
B
It will stay the same.
C
We need to know the area of the plates to answer this question.
D
It will increase.
Verified step by step guidance1
Understand the concept of a parallel plate capacitor: A parallel plate capacitor consists of two conductive plates separated by an insulating material (dielectric). When connected to a battery, it stores electrical energy by maintaining a potential difference between the plates.
Recall the formula for capacitance of a parallel plate capacitor: The capacitance \( C \) is given by \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric, \( A \) is the area of the plates, and \( d \) is the separation between the plates.
Consider what happens when the battery is removed: Once the battery is disconnected, the charge \( Q \) on the plates remains constant because there is no external circuit to allow charge to flow away.
Analyze the effect of increasing plate separation: Increasing the separation \( d \) reduces the capacitance \( C \) since \( C \) is inversely proportional to \( d \). The relationship between charge, capacitance, and potential difference is given by \( V = \frac{Q}{C} \).
Conclude the effect on potential difference: As \( d \) increases, \( C \) decreases, and since \( Q \) remains constant, the potential difference \( V \) must increase according to \( V = \frac{Q}{C} \). Therefore, the potential difference between the plates increases.
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