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A parallel-plate capacitor has a plate area of 100c * m ^ 2 and a plate separation of 2.0 cm. has been charged up to 3000 V by a battery. Now, (i) after disconnecting the battery, (ii) keeping the battery connected, the distance between the plates is increased to 5.0 cm intensity of electric field between the plates and the energy of the capacitor. ( epsilon_{0} = 8.85 * 10 ^ - 12 * F * m ^ - 1
Asked by khankaifi178 | 08 Jan, 2024, 10:12: PM

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

Answered by Thiyagarajan K | 09 Jan, 2024, 02:43: PM

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