6. An isolated capacitor of unknown capacitance has been charged to a potential

Question

6. An isolated capacitor of unknown capacitance has been charged to a potential difference of 150 V. When the charged capacitor is disconnected from the battery and then connected in parallel to an uncharged 10.0 µF capacitor, the voltage across the combination is measured to be 40.0 V. Calculate the unknown capacitance.
µF

8.A tiny sphere of mass 8.50 µg and charge −2.80 nC is initially at a distance of 1.64 µm from a fixed charge of +7.97 nC.

(a) If the 8.50- µg sphere is released from rest, find its kinetic energy when it is 0.500 µm from the fixed charge. ________J

(b) If the 8.50- µg sphere is released from rest, find its speed when it is 0.500 µm from the fixed charge.

Explanation / Answer

1) First find the capacitance: Capacitance = (the dielectric constant) )(esub o) Area/(the distance between the plates. C = k e A / d C = 1 x 8.85E-12 x .02 / 5E-6 = 35.4E-9 Farads Then find the energy: Energy = 1/2 Cv^2 = .5 x 35.4E-9 x 144 =2.55E-6J 2) Since the capacitors are in parallel, they have the same voltage. The second capacitor took 70 % of the first capacitors voltage, and therefore took 70% of the charge. The charge on C2 is 30v x 10uF = 300 uC So 300uC = .7 x origional charge on C1, = 428.6uC Since charge = v x C, C= q/v,C= 428.6/100 = 4.3 uF

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