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Two parallel-plate capacitors, 6.0 mF each, are connected in parallel to a 10 V battery. One of the capacitors is then squeezed so that its
Question
Two parallel-plate capacitors, 6.0 mF each, are connected in parallel to a 10 V battery. One of the capacitors is then squeezed so that its plate separation is 50.0% of its initial value. Because of the squeezing, (a) how much additional charge is transferred to the ca- pacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?
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Physics
5 years
2021-08-22T12:34:06+00:00
2021-08-22T12:34:06+00:00 1 Answers
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Answers ( )
The additional charge transferred to the capacitors by the battery is 60 μC.
The increase in the total charge stored on the capacitors is 60 μC.
Explanation:
C = εоA / d
If the separation is halved, then the capacitance will be doubled and according to the equation q = CV, the charge will be doubled too.
q = CV
= (6e – 6)
(10)
= 60 μC
Final charge of the capacitor:
q = (2C)V
= (2
6e – 6)
(10)
= 120 μC
additional charge transmitted is:
q’ = 120 – 60
= 60 μC
= (6 + 6)
(10)
= 120 μF
final total charge:
= (2
6 + 6)
(10)
= 180 μF
Increase in the charge:
q’ = 180 – 120
= 60 μC