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How many electrons must be added to one plateand removed from the other so as to store 25Jof energy in 5nF parallel plate capacitor.4. |
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Answer» Since q^2/2c=E. as we know value of all. q=ne . E=25J. C=5*10^-9f by Putting all values.. q=sqrt(2cE) ne=sqrt(25*2*5*10^-9) ne=5*10^-4.n=5*10^-4/(1.6*10^-19)n=5*10^15/1.6n=3.12*10^15.Now You Have No of electrons. |
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