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a charge of +2x10 ^-8 C …

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a charge of +2x10 ^-8 C is placed on the positive plate and a charge of -1x10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2x10^-3 x10^-6 F . calculate the potential difference developed between the plates

    • Posted by Nitin Jangid 5 years, 4 months ago

    CBSE > Class 12 > Physics
    • 1 answers

    Abhinay Yadav 5 years, 4 months ago

    F=kq1q2/r²

    1Thank You

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