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The circuit looks like the following going clockwise: battery, switch, capacitor.
Scenario 1: The capacitor is allowed to charge and then the switch is opened. It is placed in a highly conductive environment.
Scenario 2: The circuit switch stays closed and it is placed in a highly conductive environment.
What happens to charge of the capacitor in each of these scenarios?
My understanding: When the circuit is open, the capacitor is initially fully charged. A highly conductive environment will leech off some of the capacitors stored charge. I'm just not sure what happens when the battery is connected vs disconnected.
I would guess that if the circuit is closed and in a highly conductive environment, then some of the capacitor's stored charge will be leeched by the environment BUT the closed circuit battery will continually replace the leeched charge. This is incorrect. Why??
Scenario 1: The capacitor is allowed to charge and then the switch is opened. It is placed in a highly conductive environment.
Scenario 2: The circuit switch stays closed and it is placed in a highly conductive environment.
What happens to charge of the capacitor in each of these scenarios?
My understanding: When the circuit is open, the capacitor is initially fully charged. A highly conductive environment will leech off some of the capacitors stored charge. I'm just not sure what happens when the battery is connected vs disconnected.
I would guess that if the circuit is closed and in a highly conductive environment, then some of the capacitor's stored charge will be leeched by the environment BUT the closed circuit battery will continually replace the leeched charge. This is incorrect. Why??