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