Inserting a dielectric changes the electric field between capacitor plates and the capacitance by what factors (κ = ε_r)?

Inserting a dielectric changes the electric field between capacitor plates and the capacitance by what factors (κ = ε_r)?

• A. E decreases by κ; C increases by κ
• B. E increases by κ; C decreases by κ
• C. E stays same; C increases by κ
• D. E decreases by κ; C stays same

Answer: (A)

Inserting a dielectric raises the permittivity between the plates, which increases the capacitor’s ability to store charge. For a parallel-plate capacitor, the capacitance is C = ε A / d. Replacing ε0 with κ ε0 makes C increase by a factor of κ, so the capacitance becomes κ times larger. If the charge on the plates is held fixed (a common scenario for this question), the electric field between the plates is E = Q/(ε A). With the dielectric, ε becomes κ ε0, so E = Q/(κ ε0 A) = E0/κ. The field strength therefore decreases by a factor of κ. Note: if the capacitor were connected to a constant-voltage source instead, the field would stay the same (E = V/d) and the charge would increase by κ, but the setup implied here is fixed charge, leading to E decreasing by κ and C increasing by κ.

Inserting a dielectric raises the permittivity between the plates, which increases the capacitor’s ability to store charge. For a parallel-plate capacitor, the capacitance is C = ε A / d. Replacing ε0 with κ ε0 makes C increase by a factor of κ, so the capacitance becomes κ times larger.

If the charge on the plates is held fixed (a common scenario for this question), the electric field between the plates is E = Q/(ε A). With the dielectric, ε becomes κ ε0, so E = Q/(κ ε0 A) = E0/κ. The field strength therefore decreases by a factor of κ.

Note: if the capacitor were connected to a constant-voltage source instead, the field would stay the same (E = V/d) and the charge would increase by κ, but the setup implied here is fixed charge, leading to E decreasing by κ and C increasing by κ.