A parallel-plate capacitor connected to a battery becomes fully charged. After the capacitor from the battery is disconnected, the separation between the plates of the capacitor is doubled in such a way that no charge leaks off. How is the energy stored in the capacitor affected? A.) The energy stored in the capacitor quadruples its original value. B.) The energy stored in the capacitor remains constant. C.)The energy stored in the capacitor doubles its original value. D.)The energy stored in the capacitor is decreased to one-fourth of its original value. E.)The energy stored in the capacitor is decreased to one-half of its original value

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kollybaba55 kollybaba55 · Answer 1 · 2020-03-14T01:35:01+01:00

Answer:

C.)The energy stored in the capacitor doubles its original value.

Explanation:

The capacitance of a capacitor is given by:

C₁ = ∈A/d...............................(1)

If the distance between the plates is doubled

C₂ = ∈A/2d.............................(2a)

From equations (1) and (2), the relationship between C₁ and C₂ is:

C₂ = C₁/2....................................(2b)

The original Energy stored in the capacitor is given by :

E₁ = Q²/2C₁...............................(3)

On doubling the separation between the capacitance plates

E₂ = Q²/2C₂...............................(4a)

E₂ = Q²/2 * 1/C₂........................(4b)

Putting equation (2b) into (4b)

E₂ = Q²/2 * 2/C₁

E₂ = Q²/C₁.....................................(5)

Comparing equations (3) and (5)

E₂ = 2E₁

Therefore, the energy stored in the capacitor doubles its original value