Question

how many electrons must be added to one plate and removed from the other so as
to store 25.0J of energy in a 5.0nF parallel plate capacitor?
and how would you modify this capacitor so that it can store 50.0 J of energy without
changing the charge on its plates?​

Answer 1

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*10^15/1.6

n=3.12*10^15.

Now You Have No of electrons.