A parallel-plate capacitor with air between the plates is connected to a battery

Question

A parallel-plate capacitor with air between the plates is connected to a battery. The following steps are then carried out, but you get to decide in what order they are done.
Step A – Disconnect the battery.
Step B – Reduce the distance between the plates by a factor of 2.
Step C – Increase the distance between the plates by a factor of 3.

Let's say you now compare the following two procedures, starting from the same initial configuration with the capacitor connected to the battery.
Procedure 1: First do step B, then C, then A.
Procedure 2: First do step C, then A, then B.

What is the ratio of the potential difference across the capacitor after procedure 1 to the potential difference across the capacitor after procedure 2?

V1 /V2 =

Explanation / Answer

V1 / V2 = ?

Let the initial distance between the plates of the capacitor be d.

After the procedure 1 potential difference between the plates = V

After the procedure 2 potential difference between the plates = V/2

This is because the capacitance is increased by 2 when the distance is decreased by 2.

V1 / V2 = 2

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