Two parallel-plate capacitors, 5.6µF each, are connected in parallel to a 18 V b

Question

Two parallel-plate capacitors, 5.6µF each, are connected in parallel to a 18 V battery. One of the capacitors is then squeezedso 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, due to the squeezing?
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, due to the squeezing?
2 µC

Explanation / Answer

   Capacitance of one parallel plate capacitor C1 = o A / d = 5.6F          because ofsqueezing separation between two plates is halved       so C1 = 2 ( o A / d ) =11.2 F      Capacitance of another parallel platecapacitor C2 = o A /d = 5.6 F These two are connected in parallel               C p = C 1 + C2       = 11.2 + 5.6    = 16.8 F     a) C 1 = Q1 /V Q1 = C1 V    = 11.2 F * 18 V    = 0.2 m C    b ) Q = C p V    = 16.8 *18    = 0.302 m C

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