Soal Evaluasi Listrik Statis

C1 = 10 µF C 2 = 2 µF C 3 = 6 µF C 4 = 20 µF C 5 = 15 µf C 6 = 5µF

C P = C 2 + C3 = 2 + 6 = 8µF C P ' = C 4 + C 5 + C 6 = 20 +15 + 5 = 40 µF

Tegangan masing-masing kapasitor : 1 1 1 1 = + + CT C1 C P C P ' 1 1 1 4 5 1 10 + + = + + = 10 8 40 40 40 40 40 40 CT = = 4 µF 10 =

QT = VT .CT = 20 .4 QT = Q1 = QP = Q p ' = 80 µC

V1 =

Q1 80 µC = = 8ν C1 10 µF

VP =

QP 80 µC = CP 8µF

V P = V2 = V3 = 10ν

VP ' =

Q P ' 80 µC = = 2ν C P ' 40 µF

VP ' = V4 = V5 = V6 = 2ν

Muatan yang disimpan setiap kapasitor : Q1 = 80 µC Q2 =C 2 .V2 = 2 µF .10ν Q2 = 20 µC Q3 = C 3 .V3 = 6 µF .10ν Q3 = 60 µC Q4 = C 4 .V4 = 20 µF .2ν Q4 = 40 µC Q5 = C 5 .V5 =15 µF .2ν Q5 = 30 µC Q6 = C 6 .V6 = 5µF .2ν Q6 =10 µC

Energi Tiap Kapasitor :

Wn =

1 Qn .V n 2

W1 =

1 .8 x10 −5.8 = 3,2 x10 −4 J 2

W2 =

1 .2 x10 −5.10 =1x10 −4 J 2

W3 =

1 .6 x10 −5.10 = 3 x10 −4 J 2

W4 =

1 .4 x10 −5.2 = 4 x10 −5 J 2

W5 =

1 .3 x10 −5.2 = 3 x10 −5 J 2

W6 =

1 .1x10 −5.2 =1x10 −5 J 2