A capacitor is charged and then disconnected from a battery. A dielectric slab is then inserted between the plates. What happens to the voltage across the capacitor?
AVoltage increases because the dielectric stores extra energy
BVoltage decreases because the dielectric's induced field partially cancels the field from the plates
CVoltage stays the same because charge is conserved and the battery is disconnected
DVoltage increases by a factor of κ, matching the increase in capacitance
When the battery is disconnected, the charge Q on the plates is fixed. Inserting a dielectric polarizes the material, creating bound surface charges that produce an electric field opposing the original field. The total field inside decreases by a factor of κ, so the voltage V = Ed also decreases by κ. Since C = Q/V and Q is fixed, capacitance increases by κ (C = κC₀). This is the key chain: polarization → opposing field → reduced field → reduced voltage → increased capacitance. Option C is the trap: charge is conserved, but voltage is not — the field (and thus voltage) drops when the dielectric is inserted.
Question 2 Multiple Choice
Water has a dielectric constant κ ≈ 80, while polyethylene has κ ≈ 2.3. What accounts for water's much larger value?
AWater molecules are larger and displace more charge when the field is applied
BWater has permanent molecular dipoles that physically align with the applied field, in addition to induced dipoles
CWater conducts electricity, so charges can flow to the surfaces and amplify the field
DWater has a lower density, so polarization is more efficient per unit volume
Water is a polar molecule with a permanent electric dipole moment. When an external field is applied, those permanent dipoles tend to align with the field (thermal energy partially randomizes this alignment), contributing strongly to the polarization P. Polyethylene is non-polar and relies only on induced dipoles — the electron clouds are slightly displaced by the field. Permanent dipole alignment is much larger in magnitude than pure induction, giving water its very high κ. Water is an insulator (pure water), not a conductor, ruling out option C.
Question 3 True / False
Inside a polarized dielectric, the electric field from the bound surface charges points in the opposite direction to the applied external field.
TTrue
FFalse
Answer: True
When an external field E₀ is applied, positive charges in each atom shift slightly in the field direction and negative charges shift opposite. At the material's surfaces, this creates positive bound charges on the face toward which positive charges shifted, and negative bound charges on the opposite face. The bound surface charges then produce their own electric field E_induced that points from positive to negative — i.e., opposite to E₀. The total field inside is E = E₀ − E_induced < E₀. The dielectric partially screens the applied field.
Question 4 True / False
In a uniformly polarized dielectric, the dipoles throughout the interior cancel each other, leaving net bound charges only on the surfaces.
TTrue
FFalse
Answer: True
In a uniform polarization, every interior dipole's positive end is directly adjacent to the negative end of the next dipole. Interior charges cancel in pairs, leaving no net bound volume charge. Only at the surfaces — where the chain of dipoles terminates — are there uncompensated charges. The surface bound charge density is σ_b = P·n̂ (polarization dotted with the outward normal). This is why the macroscopic effect of a dielectric is captured entirely by its surface charges in the uniform case.
Question 5 Short Answer
Explain why inserting a dielectric into a capacitor allows it to store more charge at the same voltage.
Think about your answer, then reveal below.
Model answer: The dielectric polarizes when the field is applied, creating bound surface charges that oppose the external field. This reduces the electric field between the plates. For the same voltage (V = Ed), a weaker field means the plates must hold more charge to maintain that voltage — or equivalently, the same charge produces less voltage. Either way, C = Q/V increases by the factor κ. The dielectric effectively allows more charge to accumulate on the plates for a given potential difference.
The mechanistic chain is: applied field → dipole induction → bound surface charges → opposing field → reduced net field → reduced V for same Q → increased C = Q/V. This is why dielectrics are used in practical capacitors: they increase capacitance in a fixed volume and also prevent the plates from touching (acting as a physical spacer), allowing thinner gaps and even higher capacitance.