How many 0.25-pF capacitors must be connected in parallel in order to store 1.2µC of charge when connected to?

How many 0.25-pF capacitors must be connected in parallel in order to store 1.2µC of charge when connected to a battery providing a potential difference of 10V?

1 Answer

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  • Colin
    Lv 7
    6 years ago
    Best Answer

    Q = C*V

    C = Q/V = (1.2*10^-6/10) = 1.2*10^-7

    Capacitors in parallel add, so if we need n capacitors

    C = 1.2*10^-7 = n*0.25*10^-12

    n = 1.2*10^-7/(0.25*10^-12)

    = 4.8*10^5 = 480,000 <<<

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