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What is the difference between these two circuits from the point of view of final charge on the capacitor plates?

air gap capacitor linked to power supply with and without ground on positive side

With my understanding, circuit 'A' will quickly accumulate negative charge on the lower plate and an equal amount of positive charge on the upper plate.

In circuit 'B' I am less clear on what the difference would be given the power supply is still connected. If the net charge on the plates is the same, are there any other expected differences?

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    $\begingroup$ physics.stackexchange.com/q/279351 $\endgroup$
    – user65081
    Jan 4 '20 at 23:30
  • $\begingroup$ That seems to imply there would be no difference between A and B $\endgroup$
    – jamesj629
    Jan 5 '20 at 13:19
  • $\begingroup$ Something must change. The electric fields created by the battery must change to reflect the fact that the potential relative to infinity has changed. A charge redistribution somewhere? $\endgroup$
    – user65081
    Jan 5 '20 at 15:27
  • $\begingroup$ Let's say the battery is 9 volts. So for the battery in circuit B, the positive side is now referenced as "zero", so the negative terminal is -9V relative to the positive terminal/ground. Does that mean negative charge from ground has gone to the top plate (and positive terminal), and an equal amount of charge from the negative terminal flows to the bottom plate, keeping a 9V pd? $\endgroup$
    – jamesj629
    Jan 5 '20 at 19:54
  • $\begingroup$ It is the potential difference across the plates that determines the charge--not the potential relative to infinity. Connecting the positive plate to ground will not cause a current (dQ/dt) to flow since it does not effect to potential difference. $\endgroup$
    – user45664
    Jan 6 '20 at 0:30
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The final voltage across the capacitors would be the same. So the final charges would be the same. The only difference would be that the positive terminal of the voltage source in circuit B would be referenced to ground. Whereas the voltage source in circuit A would be 'floating'.

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The potential of the positive side for the capacitor B is always zero, because it is connected to the earth.

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