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summary Capacitance Parallel plates, coaxial cables, Earth Series and parallel combinations Energy in a capacitor Dielectrics Dielectric strength
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Series combinations reduce the capacitance
Series combinations reduce the capacitance. Equal C reduce by the number involved. In parallel the capacitance increases. A basket of 4 capacitors, each of C = 6 nF. How can you arrange them to get 1.5 nF 2 nF g) 2.4 nF 3 nF h) 3.6 nF 4 nF i) 4.5 nF j) 6 nF 12 nF k) 18 nF 24 nF
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clicker All C’s are 8.00 nF. The battery is 12 V.
What is the equivalent capacitance? 4 nF 6 nF 8 nF 10 nF 12 nF
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All C’s are 8.00 nF. The battery is 12 V.
What is the equivalent capacitance? C12 = 4 nF C123 = 12 nF Q123 = C123 x V = 144 nC Q3 = C3 x V = 96 nC Q12 = C12 x V = 48 nC U123 = ½ C123V2 = ½ x 12x10-9 x122 = 864 nJ U1 = ½ C1V12 = ½ x 8x10-9 x62 = 144 nJ = U2 U3 = ½ x 8x10-9 x122 = 576 nF C3 stores most energy, also the highest electric field and most charge, the most stressed part of the circuit.
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Circuits All capacitors being the same, rank the equivalent capacitances of the four circuits.
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When the switch S is closed, how much charge flows through point P
C1 = C3 = 8.00 μF, C2 = C4 = 6.00 μF, V = 12V When the switch S is closed, how much charge flows through point P C µF, q = 28.8 µC C2 C24 = 12 µF C1234 = 3 µF q =36 µC Δq = 7.2 µC
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q -q q' -q' V V' ( )
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q -q q' -q' V V' ( )
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If the areas are A1 and A-A1.
Effect of a dielectric : C κC C µF, q = 28.8 µC C2 C24 = 12 µF C1234 = 3 µF q =36 µC
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The force on a filling dielectric as it is inserted between the parallel plates of a capacitor.
x L With the battery connected, U1 = ½CV2 With the battery disconnected, U2 = Q2/2C With the battery connected, since x is increasing downwards, a negative force is upwards, pushing the dielectric away. With the battery disconnected, the force is positive and pointed downwards, pulling in the dielectric. The force is proportional to (κ-1) and inversely to L.
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A question What is the equivalent capacitance between the points A and B? 1 μF B. 2 μF 4 μF 10μF None of these A B What would a 10V battery do, i.e. how much charge will it provide, when it is connected across A and B? 40 μC
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Question A parallel-plate capacitor has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about: 5N 0. 9N 1 x104 N 9 x 105 N The electric field = σ/2εo why?
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Question A parallel-plate capacitor has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about: A. 0 B. 5N C. 9N D. 1 x104 N E. 9 x 105 N
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A question Each of the four capacitors shown is 500 μF. The voltmeter reads 1000V. The magnitude of the charge, in coulombs, on each capacitor plate is: A. 0.2 B. 0.5 C. 20 D. 50 E. none of these
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HITT A parallel-plate capacitor has a plate area of 0.2m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 106 V/m between the plates, the magnitude of the charge on each plate should be: A. 8.9 x 10-7 C B. 1.8 x 10-6 C C. 3.5 x 10-6 C D. 7.1 x 10-6 C E. 1.4 x 10-5 C c
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