Energy Ratio in Capacitor on Dielectric Slab Insertion – Rankers Physics
Topic: Capacitors
Subtopic: Capacitor With Dielectrics

Energy Ratio in Capacitor on Dielectric Slab Insertion

A parallel plate capacitor has plates with area \(A\) and separation \(d\). A battery charges the plates to a potential difference \(V_0\). The battery is then disconnected and a dielectric slab of thickness \(d\) is introduced. The ratio of energy stored in the capacitor before and after slab is introduced, is:
\(K\)
\(\frac{1}{K}\)
\(\frac{A}{d^2K}\)
\(\frac{d^2K}{A}\)

Solution:

When the battery is disconnected, the charge \(Q\) remains constant. The initial energy is \(U_i = \frac{Q^2}{2C_0}\). After inserting the dielectric of constant \(K\), the capacitance becomes \(C = K C_0\) and the final energy is \(U_f = \frac{Q^2}{2K C_0} = \frac{U_i}{K}\). Therefore, the ratio \(U_i / U_f = K\).

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