A capacitor is charged to voltage V and then disconnected from the battery. A dielectric with κᵣ = 4 is inserted between the plates. What happens to the electric field inside the capacitor?
AIt increases by a factor of 4, because the dielectric amplifies the field
BIt decreases to E/4, because the dielectric's bound charges oppose the applied field
CIt stays the same — the dielectric only affects capacitance, not the field inside
DIt drops to zero, because the dielectric perfectly cancels the applied field
Inserting a dielectric reduces the field inside by a factor of κᵣ: E_inside = E_free/κᵣ. The physical mechanism is that the bound surface charges create a field opposing the original field, partially canceling it. With κᵣ = 4, the internal field becomes E/4. Note: if the capacitor were connected to a battery (fixed voltage), the field would remain the same but charge would increase. Here, disconnected from the battery means charge Q is fixed, so reducing the field means voltage across the capacitor also drops by a factor of 4.
Question 2 Multiple Choice
Why does water (κᵣ ≈ 80) reduce the electric field inside it far more than most plastics (κᵣ ≈ 2–4)?
AWater is a conductor, so it redistributes charge to cancel the internal field
BWater molecules are permanent electric dipoles that align strongly with applied fields, producing large opposing bound charge sheets
CWater's higher density packs more molecules per unit volume, multiplying the bound charge effect
DWater is transparent to electric fields, allowing them to pass through with reduced strength
Water molecules have a permanent dipole moment (they are polar) due to the asymmetric arrangement of hydrogen and oxygen. In an applied field, these permanent dipoles align very effectively, creating a large macroscopic polarization P. This polarization produces bound surface charges whose opposing field is much stronger than what non-polar molecules (like most plastics) can produce through induced dipoles alone. The result: E_inside = E_free/80 for water. Water is an insulator in this context — its large κᵣ comes from polarization, not conduction.
Question 3 True / False
The electric field inside a dielectric is stronger than the applied external field, because the aligned dipoles add their fields to the original.
TTrue
FFalse
Answer: False
The aligned dipoles create bound surface charges — positive on the face toward the negative plate, negative on the face toward the positive plate. These surface charges produce a field pointing opposite to the applied field, partially canceling it. The net internal field is always weaker than the applied field: E_inside = E_free/κᵣ, where κᵣ ≥ 1. The polarization reduces the field; it does not amplify it.
Question 4 True / False
If a dielectric is inserted between capacitor plates while the voltage is held constant by a connected battery, the capacitor stores more charge than without the dielectric.
TTrue
FFalse
Answer: True
Capacitance increases by a factor of κᵣ when a dielectric fills the gap: C = κᵣε₀A/d. With fixed voltage V, the charge stored is Q = CV = κᵣε₀AV/d — a factor of κᵣ more than without the dielectric. The battery supplies additional charge to maintain the same voltage while the dielectric's bound charges effectively increase the capacitance. This is the primary practical reason dielectrics are used in capacitors.
Question 5 Short Answer
Explain the microscopic mechanism by which a dielectric reduces the electric field inside it. What role do bound charges play, and where do they appear?
Think about your answer, then reveal below.
Model answer: When an external field is applied, the microscopic dipoles in the dielectric (permanent or induced) align with the field. Inside the bulk, neighboring dipoles cancel — the positive end of one dipole sits next to the negative end of the next. But at the two surfaces perpendicular to the field, this cancellation breaks down: a net sheet of positive bound charge appears on one face and negative bound charge on the other. These surface charge sheets create their own electric field pointing opposite to the applied field, reducing the total field inside to E_inside = E_applied/κᵣ.
The key is that bound charges appear only at surfaces (in a uniform dielectric), not in the bulk. This is because internal dipoles cancel each other. The opposing field from the surface charges is what produces the screening effect. Materials with larger κᵣ align their dipoles more effectively, producing stronger bound surface charge sheets and more screening. This mechanism — macroscopic screening by microscopic dipole alignment — is captured by replacing ε₀ with ε₀κᵣ in Gauss's law, eliminating the need to explicitly track bound charges.