A 10.0 μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is ful

step 1 In the given problem, the capacitance of the parallel plate capacitor is given as C is equal to

A 10.0 μF parallel-plate capacitor is connected to a 12.0 V...

A $10.0 \mu \mathrm{F}$ parallel-plate capacitor is connected to a $12.0 \mathrm{~V}$ battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation was doubled; (ii) the radius of each plate was doubled, but the separation between the plates was unchanged?

Key Concepts

Impact of Geometric Alterations on Capacitance

Modifying the physical dimensions of a capacitor directly affects its capacitance. Doubling the plate separation (d) results in halving the capacitance, thereby doubling the voltage for a fixed charge. Conversely, doubling the radius increases the plate area, which increases the capacitance (typically by a factor proportional to the area increase), and thus reduces the voltage for a given constant charge.

Capacitor Charging and Isolation

When a capacitor is initially connected to a battery, it charges up until the voltage across its plates equals the battery's voltage. Once the battery is disconnected, the total charge on the capacitor remains fixed even if the physical dimensions of the capacitor change. This condition of constant charge is essential when analyzing subsequent changes in capacitance and voltage.

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by a fixed distance with an insulating medium between them. Its capacitance is determined by the formula C = ??A/d, where A is the area of a plate and d is the separation between the plates. This relationship shows that a larger plate area increases capacitance, whereas a larger separation decreases it.

Voltage and Capacitance Relationship

For an isolated capacitor with fixed charge Q, the voltage across it is given by V = Q/C. Changes to the capacitor’s geometry, which alter the capacitance, will inversely affect the voltage. Thus, if the capacitance decreases, the voltage increases, and vice versa.

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