CBSE Class 12-science Answered
Let A = 100 cm2 be the area of capacitor plates. Initial capacitance C of capacitor , when separation between plates is d = 2 cm is
C = 4.427 × 10-12 F
Let V = 3000 V be the potential difference between capacitor plates .
Charge on the capacitor , Q = C V = 4.427 × 10-12 × 3000 C = 1.328 × 10-8 C
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Part(i) after battery is disconnected
if plate separation is increased to d1 = 5 cm , then changed capacitance C1 is
C1 = 4.427 × 10-12 × (2/5) F = 1.771 × 10-12 F
Potential difference across plates , V1 = Q / C1 = ( 1.328 × 10-8 C / 1.771 × 10-12 F ) = 7500 V
Electric field E = V/d1 = 7500/( 5 × 10-2 ) V/m = 1.5 × 105 V/m
Energy stored in capacitor , U = (1/2) C V2 = 0.5 × 1.771 × 10-12 × ( 7500 )2 = 5 × 10-5 J
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Part (ii) Battery is not disconnected
As done in previous part , if plate separation is increased to d1 = 5 cm , then changed capacitance C1 is 1.771 × 10-12 F .
Potential difference across capacitor plates is same as the potetial difference across battery i.e. 3000 V
Electric Field E = V/d1 = 3000 / ( 5 × 10-2 ) V/m = 6 × 104 V/m
Energy stored in capacitor , U = (1/2) C V2 = 0.5 × 1.771 × 10-12 × ( 3000 )2 = 7.970 × 10-6 J