I have the right answer but dont know how i got it. Two parallel-plate capacitor

Question

I have the right answer but dont know how i got it. Two parallel-plate capacitors, 2.0 µF each, are connected in series to a9.0 V battery. One of the capacitorsis then squeezed so that its plate separation is halved. (a) Because of the squeezing, how muchadditional charge is transferred to the capacitors by thebattery?
3 µC

(b) What is the increase in the total charge stored on thecapacitors? 6 µC I have the right answer but dont know how i got it. Two parallel-plate capacitors, 2.0 µF each, are connected in series to a9.0 V battery. One of the capacitorsis then squeezed so that its plate separation is halved. (a) Because of the squeezing, how muchadditional charge is transferred to the capacitors by thebattery?
3 µC

(b) What is the increase in the total charge stored on thecapacitors? 6 µC

Explanation / Answer

initially, total capacitance is    2 *2 /(2 + 2) =    1 uF . So total charge stored is    1 * 9 =   9 uC . When one cap is squeezed, its capacitance doubles to 4. Nowthe total capacitance is .      2 * 4 / (2 + 4) =  8/6 = 4/3 . And the total charge stored willbe      capacitance * voltage = 4/3 * 9   = 12 uC . So the additionalcharge is     final - initial   = 12 - 9  =   3 uC . Each capacitor gains anadditional   3 uC,   so the total additionalcharge stored is 6 uC

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