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

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

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.

Circuits All capacitors being the same, rank the equivalent capacitances of the four circuits.

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 C123 2.4 µF, q = 28.8 µC C2 C24 = 12 µF C1234 = 3 µF q =36 µC Δq = 7.2 µC

q -q q' -q' V V' (25 - 15)

q -q q' -q' V V' (25 - 16)

If the areas are A1 and A-A1. Effect of a dielectric : C  κC C123 2.4 µF, q = 28.8 µC C2 C24 = 12 µF C1234 = 3 µF q =36 µC

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.

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

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?

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

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

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