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Area of each plate of the parallel plate capacitor `A=6xx10^(-3)m^(2)` <br> Distance between the plates , d=3mm=`3xx10^(-3) m ` <br> supply voltage, v=100 V <br> capacitance C of a parallel plate capacitor is given by, <br> `C=(in_(0)A)/d` <br> Where, `in_(0)` =permittivity of free space <br> `=8.854xx10^(-12) N^(-1)m^(-2)C^(-2)` <br> `:. C=(8.854xx10^(-12)xx6xx10^(-3))/(3xx10^(-3))` <br> `17.71xx10^(-12)f` <br> `=17.71 pF` <br> Potential V is related with the charge q and capacitance C as `V=q/C` <br> `:. q=VC` <br> `=100xx17.71xx10^(-12)` <br> `=1.771xx10^(-9)C` <br> Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is `1.771xx10^(-9)C` Related Video

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