Effect of inserting a dielectric slab on stored energy in a parallel-plate capacitor A parallel-plate capacitor is connected to a constant-voltage source. When a dielectric slab is fully inserted between the plates, the energy stored in the capacitor increases to five times its original value. What is the dielectric constant (relative permittivity) of the material?

Introduction / Context:
Dielectrics increase the capacitance of a parallel-plate capacitor by a factor equal to their relative permittivity. Whether the stored energy increases or decreases upon inserting a dielectric depends on whether the capacitor is isolated (constant charge) or connected to a source (constant voltage). This distinction is frequently examined in circuits and electromagnetics courses.

Given Data / Assumptions:

• Capacitor remains connected to a constant voltage V.
• Energy increases to five times the original value after insertion.
• Dielectric completely fills the space between plates (no fringing effects considered).

Concept / Approach:

Original energy: U0 = 0.5 * C0 * V^2. With a dielectric: C = κ * C0 where κ is the dielectric constant (relative permittivity). At constant voltage, new energy is U = 0.5 * (κ C0) * V^2 = κ * U0. An increase by a factor of five implies κ = 5. (Note: if the capacitor were isolated, U would become U0 / κ.)

Step-by-Step Solution:

Verification / Alternative check:

Dimensional and physical checks: In constant-voltage condition, source supplies additional charge to maintain V as C increases, hence energy rises by κ. This matches observed behavior in lab demonstrations.

Why Other Options Are Wrong:

1/5 and 5/2 do not produce fivefold energy increase under constant V; 25 would cause a 25× increase; “Cannot be determined” is wrong because the operating condition is specified.

Common Pitfalls:

Confusing constant-charge and constant-voltage cases; forgetting that energy changes sign depending on the constraint.

Final Answer:

5