Answer
Total charge on 1500"\\mu F" capacitor
"Q=CV\\\\=0.0015\\times100\\\\Q=0.15columb"
When supply is disconnected and another capacitor is connected then
Voltage across each capacitor
"V_1=\\frac{0.15}{0.0015+0.0010}=60volts"
(ii) initial energy stored
"U_i=\\frac{1}{2}CV^2=\\frac{1}{2}(0.0015)(100)^2\\\\U_i=7.5J"
Final energy
"U_f=\\frac{1}{2}C_{eq}V_1^2\\\\=\\frac{1}{2}(0.0015+0.0010)(60)^2=4.5J"
Loss in energy ="U_i-U_f=7.5-4.5\\\\=3J"
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