Two parallel-plate capacitors, 2.0µF each, are connected in series to a 9.0 V ba

Question

Two parallel-plate capacitors, 2.0µF each, are connected in series to a 9.0 V battery. One of the capacitors is thensqueezed so that its plate separation is halved. (a) Because of the squeezing, how muchadditional charge is transferred to the capacitors by thebattery?
1 µC

(b) What is the increase in the total charge stored on thecapacitors?
2 µC (a) Because of the squeezing, how muchadditional charge is transferred to the capacitors by thebattery?
1 µC

(b) What is the increase in the total charge stored on thecapacitors?
2 µC

Explanation / Answer

total capacitance initially is    2.0 /2   =   1.0 uF    so initialtotal charge is   1.0 * 9.0 = 9.0 uC . total capacitance finally is    2.0 * 4.0 /(2.0 + 4.0) =   8/3 uF    . final total charge is    (8/3) * 9 =    24.0 uC . additional charge = 24 - 9  = 15.0 uC . (b)   this seems to ask the same question... theadditional charge transferred by the battery is the same thing asthe increase in total charge stored on the capacitors.

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