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Keeping the voltage of the charging source constant, what would be the percentage change in the energy stored in a parallel plate capacitor if the separation between its plates is decreased by 10%? |
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Answer» Solution :Energy stored in the capacitor is given by `E=1/2 CV^(2) and C=(epsi_(0)A)/(d)`, where d is the separation by between the parallel plate capacitor. When d is DECREASED by 10% = 0.10 New capacitance, `C_1=(epsi_(0)A)/(0.90) =(10 epsi_(0) A)/(9)=(10)/(9)C` New energy stored, `E_(1)=1/2. (10)/(9) CV^(2)=(10)/(9) E` Change `=E_(1)-E` `"% change"=(E_(1)-E)/(E) xx 100=((10)/(9)-1) xx 100=11.1%` |
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