A parallel-plate capacitor of plate separation d is charged...

A parallel-plate capacitor of plate separation $d$ is charged to a potential difference $\Delta V_{0}$ . A dielectric slab of thickness $d$ and dielectric constant $\kappa$ is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is $U / U_{0}=\kappa$ Give a physical explanation for this increase in stored energy. (b) What happens to the charge on the capacitor? (Notice that this situation is not the same as in Example $26.5,$ in which the battery was removed from the circuit before the dielectric was introduced.)

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Key Concepts

Energy Storage in Capacitors

Energy stored in a capacitor is determined by the capacitance and the voltage across it. For a capacitor connected to a battery (constant voltage), inserting a dielectric increases the capacitance and therefore the energy stored in the capacitor, as the stored energy is proportional to the capacitance.

Effect of a Constant Voltage Source

When a battery remains connected to a capacitor, it maintains a fixed voltage across the plates. As capacitance increases through the insertion of a dielectric, the battery supplies additional charge to keep the voltage constant. This highlights the difference from an isolated capacitor, where the charge would remain fixed in the absence of an external voltage source.

Capacitance and Dielectrics

This concept explains that inserting a dielectric material between the plates of a capacitor increases its capacitance by reducing the effective electric field within the capacitor. The new capacitance is the product of the dielectric constant (?) and the original capacitance, allowing more charge to be stored at a given voltage.

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