A battery maintains a constant 100 V across a parallel-plate capacitor. A dielectric with κ = 3 is inserted between the plates. What correctly describes the new state?
AVoltage triples to 300 V because the dielectric amplifies the field
BCharge on the plates triples because capacitance increases while voltage stays fixed at 100 V
CCharge stays the same because charge is conserved
DCapacitance triples but charge stays the same, so voltage must drop to 33 V
With the battery connected, voltage is held fixed at V = 100 V. Inserting the dielectric increases capacitance from C₀ to κC₀ = 3C₀. Since Q = CV, charge triples from Q₀ to 3Q₀ — the battery pumps extra charge onto the plates. Energy stored (½CV²) also triples. This contrasts sharply with the battery-disconnected scenario: the battery-connected case is governed by fixed V, while the disconnected case is governed by fixed Q.
Question 2 Multiple Choice
A dielectric reduces the electric field inside a capacitor. How does this happen?
AMobile charges within the dielectric flow to the plates, partially neutralizing them
BThe dielectric absorbs energy from the field, converting it to heat
CMolecular dipoles in the dielectric align to create an internal field opposing the applied field
DThe dielectric increases the effective plate separation, which reduces E = V/d
A dielectric is an insulator — no charge flows through it. Instead, its molecules respond to the applied field by polarizing: either permanent dipoles rotate to align with the field, or nonpolar molecules develop induced dipoles. These aligned dipoles collectively produce an internal electric field that points opposite to the applied field, partially canceling it. The net field inside is reduced by the factor κ. This is entirely distinct from a conductor's response, where free charges do physically redistribute.
Question 3 True / False
Inserting any real dielectric material between capacitor plates always increases the capacitance, never decreases it.
TTrue
FFalse
Answer: True
The dielectric constant κ is defined as the ratio of permittivity of the material to that of free space, and κ > 1 for all real dielectric materials. Therefore C = κε₀A/d > ε₀A/d = C₀ always. This is a universal property: polarization always partially opposes the applied field, which always reduces V for fixed Q (or equivalently, allows more Q to be stored for fixed V). There is no real material with κ < 1 that would decrease capacitance; such a material would have to amplify rather than oppose the field.
Question 4 True / False
A capacitor is charged to voltage V₀ and then the battery is disconnected. Inserting a dielectric with κ = 2 increases the energy stored in the capacitor.
TTrue
FFalse
Answer: False
With the battery disconnected, charge Q is fixed. Inserting the dielectric doubles capacitance (C → 2C₀) while Q stays constant. Energy stored = Q²/(2C). Since C doubled, energy is halved: U = Q²/(2·2C₀) = U₀/2. The dielectric is actually pulled between the plates by the fringe fields, and this mechanical work done by the field accounts for the energy decrease — the dielectric gains kinetic energy as it is pulled in. Energy is conserved; it just changes form. This is the opposite of the battery-connected case, where inserting the dielectric increases stored energy.
Question 5 Short Answer
Explain why inserting a dielectric reduces the electric field inside the capacitor, and why this reduction does not involve any charge flowing through the material.
Think about your answer, then reveal below.
Model answer: The dielectric reduces E because its molecules polarize in the applied field — either rotating existing dipoles or inducing temporary ones. These aligned dipoles create their own internal electric field pointing opposite to the applied field, partially canceling it. No charge flows because the material is an insulator; the dipoles are bound within individual molecules and simply reorient. The result is a reduced net field E = E₀/κ inside the material.
The key distinction is between bound charge (dipole reorientation within molecules) and free charge (conduction). In a conductor, free electrons redistribute to cancel E completely. In a dielectric, bound charges merely shift slightly within their molecules. This partial, local response reduces E by κ but cannot eliminate it. The misconception that dielectrics work by conducting charge away is common but wrong — a dielectric that conducted charge would be a capacitor short circuit, not an insulator.