A parallel-plate capacitor is isolated with a charge ±Q on each plate. If the separation of the

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A parallel-plate capacitor is isolated with a charge ±Q on...

A parallel-plate capacitor is isolated with a charge $\pm Q$ on each plate. If the separation of the plates is halved and a dielectric (constant $K )$ is inserted in place of air, by what factor does the energy storage change? To what do you attribute the change in stored potential energy? How does the new value of the electric field between the plates compare with the original value?

Key Concepts

Interplay of Geometric and Material Changes on Stored Energy

The change in stored potential energy is a result of the combined effects of reducing the plate separation and introducing a dielectric. Halving the separation increases the capacitance by a factor of 2, while the dielectric increases it by a factor of K. This increased capacitance reduces the voltage for a given fixed charge, leading to a lower stored energy. The phenomenon highlights how both geometry and materials influence the behavior of capacitors.

Electric Field in a Capacitor with a Dielectric

The electric field between the plates of a capacitor is directly related to the surface charge density and inversely related to the permittivity of the medium. For a capacitor with constant charge, the introduction of a dielectric increases the permittivity (from ?? to K??) and thereby reduces the electric field to E = Q/(K??A). This shows that the electric field is lower by a factor of K compared to the initial configuration with air.

Capacitance of a Parallel?Plate Capacitor

The capacitance of a parallel?plate capacitor depends on the area of the plates, the separation between them, and the permittivity of the medium in the gap. When the distance between the plates is reduced, the capacitance increases because the electric field lines are brought closer together, and when a dielectric is inserted, the effective permittivity increases by a factor equal to the dielectric constant K, further increasing the capacitance.

Energy Storage in a Capacitor under Constant Charge

For an isolated capacitor where the charge remains constant, the energy stored (given by U = Q²/(2C)) is inversely proportional to the capacitance. Therefore, any increase in capacitance leads to a decrease in stored energy. In this case, halving the separation and inserting a dielectric multiplies the capacitance by 2K, reducing the energy stored by a factor of 1/(2K).

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