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Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential difference is applied across the plates? (Choose from: Increasing the area of the plates; Decreasing the area of the plates; Increasing the separation between the plates; Decreasing the separation between the plates; Inserting a material with a higher dielectric constant between the plates

User AlanK
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1 Answer

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Answer:

increasing the separation between the plates

Step-by-step explanation:

The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because

Q(charge) = CV V = VOLTAGE , c = capacitance

E = 1/2 eAV^2/ D ( ENERGY STORED )

where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference

User Stanly T
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