A dielectric with dielectric constant κ is inserted between plates of a capacitor that remains connected to a fixed voltage source. What is the effect on the electric field between the plates?

A dielectric with dielectric constant κ is inserted between plates of a capacitor that remains connected to a fixed voltage source. What is the effect on the electric field between the plates?

• A. The field decreases
• B. The field increases
• C. The field remains the same
• D. The field becomes zero

Answer: (C)

When a capacitor remains connected to a fixed voltage source, the potential difference across the plates is held constant. The electric field between the plates is set by this voltage and the plate separation, so the field is E = V/d and does not depend on the dielectric present. Inserting a dielectric increases the capacitance (C = κ ε0 A / d) and, to keep the same voltage, the battery supplies more free charge on the plates (Q = C V). The dielectric polarization does create bound charges that would oppose the field for a given free charge, but the battery compensates by increasing the free charge so that the overall field remains V/d. So the electric field between the plates stays the same.

When a capacitor remains connected to a fixed voltage source, the potential difference across the plates is held constant. The electric field between the plates is set by this voltage and the plate separation, so the field is E = V/d and does not depend on the dielectric present.

Inserting a dielectric increases the capacitance (C = κ ε0 A / d) and, to keep the same voltage, the battery supplies more free charge on the plates (Q = C V). The dielectric polarization does create bound charges that would oppose the field for a given free charge, but the battery compensates by increasing the free charge so that the overall field remains V/d.

So the electric field between the plates stays the same.