A battery is permanently connected to a parallel plate capacitor and the energy stored is x joules. When one plate is moved so that the separation of the plate is doubled, the energy now stored in the joule is:
Correct answer: C. x/2
- A. 4x
- B. 2x
- C. x/2
- D. x/4
Explanation
C = A𝛆 / d where C is capacitance, 𝛆 is the permittivity of free space, A is the area of plates, and d is the distance between them. When d is doubled, C halves.Since v remains constant throughout because the battery is permanently connected to the capacitor we use the formula E= Cv^2 / 2 where E is energy stored in a capacitor, C is capacitance, and v is the potential difference. So when C halves, E halves because C and E are directly proportional. So the answer is x/2 which is option C.
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About Capacitors
Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
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