Two parallel-plate capacitors, 6.7 ?F each, are connected in parallel to a 98 V
ID: 2280488 • Letter: T
Question
Two parallel-plate capacitors, 6.7 ?F each, are connected in parallel to a 98 V battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional charge in ?C is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?
(a)
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(b)
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(a)
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mm rad/m or m^-1 m/min mg kW m^2/s^3 m^2/s^2 A V mJ ?C F m^2 ?s Two parallel-plate capacitors, 6.7 ?F each, are connected in parallel to a 98 V battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional charge in ?C is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?Explanation / Answer
C = epsilon_o* A/ d
If distance is halved then capacitance doubles.
Initailly, Q = C V (steady state)
Q = 6.7*10^-6 * 98 = 6.566*10^-4 C
Finally , Q' = 2Q = 1313.2?C ( since capacitance has doubled)
Additional charge transferred
a) Additional charge = delta(Q) = Q' - Q = 656.6 ?C
b) increase in the total charge stored on the capacitors = delta(Q) = 656.6 ?C
Charge on capacitor 2 does not changes.
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