When a dielectric is inserted between the plates while the free charge on the plates remains fixed, how does the electric field change?

When a dielectric is inserted between the plates while the free charge on the plates remains fixed, how does the electric field change?

• A. E' = κ E
• B. E' = E
• C. E' = E/κ
• D. E' = 0

Answer: (C)

With the free charge on the plates fixed, the displacement field D is set by that free charge: D = Q/A and does not depend on the dielectric. The dielectric changes the permittivity to ε = ε0 κ, and the relation D = ε E gives E = D/ε = (Q/A) / (ε0 κ) = (Q/A)/ε0 × 1/κ. The field without the dielectric is E0 = (Q/A)/ε0, so the new field is E' = E0/κ, i.e., E' = E/κ. Physically, the dielectric becomes polarized, producing bound charges that oppose the applied field, reducing the net electric field inside. If instead the voltage were held fixed, the field would remain E = V/d, independent of κ.

With the free charge on the plates fixed, the displacement field D is set by that free charge: D = Q/A and does not depend on the dielectric. The dielectric changes the permittivity to ε = ε0 κ, and the relation D = ε E gives E = D/ε = (Q/A) / (ε0 κ) = (Q/A)/ε0 × 1/κ. The field without the dielectric is E0 = (Q/A)/ε0, so the new field is E' = E0/κ, i.e., E' = E/κ.

Physically, the dielectric becomes polarized, producing bound charges that oppose the applied field, reducing the net electric field inside. If instead the voltage were held fixed, the field would remain E = V/d, independent of κ.