# One side of capacitor plate connected to ground

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

• physics.stackexchange.com/q/279351
– user65081
Jan 4 '20 at 23:30
• That seems to imply there would be no difference between A and B Jan 5 '20 at 13:19
• 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?
– user65081
Jan 5 '20 at 15:27
• 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? Jan 5 '20 at 19:54
• 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. Jan 6 '20 at 0:30