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Two capacitors, each of capacitance 5 µF, are connected in series and then to a 100 V battery. The charge on each capacitor is:a) 250 µCb) 500 µCc) 1000 µCd) 2000 µC

Question

Two capacitors, each of capacitance 5 µF, are connected in series and then to a 100 V battery. The charge on each capacitor is:a) 250 µCb) 500 µCc) 1000 µCd) 2000 µC

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Solution

When capacitors are connected in series, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of the individual capacitances. In this case, with two capacitors each of 5 µF, the total capacitance would be:

1/C_total = 1/C1 + 1/C2 1/C_total = 1/5 + 1/5 1/C_total = 2/5 C_total = 5/2 = 2.5 µF

The charge (Q) on a capacitor is given by the product of its capacitance (C) and the voltage (V) across it. So, the charge on each capacitor would be:

Q = C_total * V Q = 2.5 µF * 100 V Q = 250 µC

So, the correct answer is a) 250 µC.

This problem has been solved

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