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How many electrons must be added to one plate and removed from the other so as to store 25.0 J of energy in a 5.0 nF parallel plate capacitor? |
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Answer» Solution :`C=5 xx 10^(-9) F, U=25J` `U=Q^(2)//2C` `Q^(2)=2UC=2 xx 25 xx 5 xx 10^(-9)` `Q=5 xx 10^(-4)C` `Q=ne` `n=Q/e=3.125 xx 10^(15)"ELECTRONS".` |
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