Tvvo identica para el pla e capacitors, ea wth capacitar e Finally, the plate se

Question

Tvvo identica para el pla e capacitors, ea wth capacitar e Finally, the plate separation in one of the capacitors is doubled. are t at ge to potential c e ence 4 . and then asco rmette rom the a ery. The are ther connecte to each i er n ata e with ates of t signo nnected (a) Find the total energy of the system of two capacitors before the plate separation is doubled Your response is off by amultiple of ten. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled (d) Reconcile the difference in the answers to parts (a) and () with the law of conservation of energy. Positive work is done by the agent pulling the plates apart. Negative work is dcne by the agent pulling the plates apart. No work is done by pulling the gent pulling the plates apart,

Explanation / Answer

part a )

Because the capacitors are connected in parallel, their voltage remains the same

U = 1/2 * CV^2 + 1/2 * CV^2

U = CV^2

given : C = 18.5 uF

V = 48 V

U = 0.0426 J = 4.26 x 10^-2 J

part b )

we know C = keo*A/d

now d = 2d for one capacitor

C' = C/2

here total charge remain same  

Q = CV

2CV = CV' + C'V'

2CV = CV' + CV'/2

V' = 4V/3 = 4 * 48/3 = 64 V

part c )

after the plate sepration is doubled

U = 1/2 * CV'^2 + 1/2 * C'V'^2

C' = C/2

U = 1/2 * CV'^2 + 1/4 * CV'^2

V' = 64 V

C = 18.5 uF = 18.5 x 10^-6 F  

U = 0.0568 J = 5.68 x 10^-2 J

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