Capacitor filled with dielectric after charging A vacuum parallel-plate capacitor is first charged and then isolated. The initial electric field between the plates is 2 × 10^4 V/m. If the space between the plates is subsequently filled completely with a dielectric material of relative permittivity εr = 10, what will be the new electric field inside the dielectric (assume the free charge on plates remains constant)?

Introduction / Context:
When a charged parallel-plate capacitor is filled with a linear dielectric after being isolated from any source, the electric field inside the dielectric changes because polarization reduces the internal field. This question tests understanding of what remains constant (free charge) versus what changes (field and potential) during dielectric insertion.

Given Data / Assumptions:

• Initial field in vacuum: E0 = 2 × 10^4 V/m.
• Relative permittivity of dielectric: εr = 10.
• Capacitor is already charged and then disconnected (free plate charge Q is constant).
• Uniform, linear, isotropic dielectric fully fills the gap.

Concept / Approach:

For a given free charge Q on the plates, inserting a dielectric increases capacitance by εr. The electric displacement D equals ε0 E in vacuum and ε0 εr E in the dielectric, but D due to free charge stays the same when Q is fixed. Therefore the electric field scales as E = E0 / εr when the dielectric fills the space after disconnection.

Step-by-Step Solution:

Verification / Alternative check:

Potential difference V = E d; since E decreases by 10 while d is unchanged, V also drops by 10, consistent with C increasing by 10 at constant Q (Q = C V). Energy reduces accordingly, with the difference appearing as work done in drawing in the dielectric.

Why Other Options Are Wrong:

• 1 × 10^4 V/m: only halves E, not correct for εr = 10.
• 2 × 10^5 V/m or 210 × 10^4 V/m: imply field increase, contrary to physics.
• 2 × 10^4 V/m: ignores dielectric effect after isolation.

Common Pitfalls:

Confusing “connected to source” (constant V) with “isolated” (constant Q). If connected to a battery, E would stay the same and Q would increase by εr; when isolated, E reduces by εr.

Final Answer:

2 × 10^3 V/m