How many electrons must be added to one plate and removed from other so as to store 25.0 J of energy in a 5.0 nF J parallel plate capacitor
Answers
Answered by
13
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*10^15/1.6
n=3.12*10^15.
Now You Have No of electrons.
q=sqrt(2cE)
ne=sqrt(25*2*5*10^-9)
ne=5*10^-4.
n=5*10^-4/(1.6*10^-19)
n=5*10^15/1.6
n=3.12*10^15.
Now You Have No of electrons.
Similar questions