1.

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.

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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