A parallel plate capacitor whose capacitance C is $$14$$ pF is charged by a battery to a potential difference $$V=12V$$ between its plates. The charging battery is now disconnected and a porcelain plate with $$k=7$$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of $$____$$ pJ. (Assume$$ no $$friction)

Question

A parallel plate capacitor whose capacitance C is $$14$$ pF is charged by a battery to a potential difference  $$V=12V$$ between its plates. The charging battery is now disconnected and a porcelain plate with $$k=7$$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of $$____$$ pJ. (Assume$$ no $$friction)

Infinity Learn · Accepted Answer

The battery is charged to a potential U the energy stored in capacitor
$U = \frac{Q^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 14 \times 12 \times 12$ $= 7 \times 144 p J$
When a dielectric plate is introduced the energy will change keeping charge constant as it is disconnected with battery
The new energy of capacitor $U^{1} = \frac{Q^{2}}{2 C^{1}} = \frac{Q^{2}}{2 C K} = \frac{U}{K} = \frac{7 \times 144}{7}$ $= 144 p J$
The change in energy will impose to the plate and it oscillates with energy of
$U - U^{1} = 7 \times 144 - 144 = 6 \times 144 = 864 p J$
[Note: Assume thickness of porcelain plate same as gap between capacitor plates]

Insertion of dielectric in capacitor

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