AP Physics C: E&M:: Conductors Capacitors Dielectrics

How is it PROVEN, using Gauss's law, that the ELECTRIC FIELD INSIDE a CONDUCTOR in ELECTROSTATIC EQUILIBRIUM must be EXACTLY ZERO, by considering what would happen if a nonzero field existed inside?

- **A)** This concept has no actual relationship between the absence of charge motion (equilibrium) and the electric field being zero inside a conductor
- **B)** This proof requires the conductor to have ZERO net charge, contradicting its actual general validity for a conductor carrying ANY net charge (positive, negative, or zero)
- **C)** The electric field inside a conductor in electrostatic equilibrium is generally NONZERO, contradicting the actual zero-field result required by equilibrium
- **D)** If a nonzero field existed inside a conductor, the conductor's FREE (mobile) charges would experience a force and ACCELERATE in response -- but 'electrostatic equilibrium' means charges are NOT moving, so the field inside MUST be zero; applying Gauss's law to any Gaussian surface drawn just inside the conductor's surface then shows the ENCLOSED charge must also be zero, forcing any NET charge to reside entirely on the OUTER surface

How does the ZERO-INTERIOR-FIELD result for a conductor, combined with E=-∇V (or E=-dV/dr), PROVE that the ENTIRE CONDUCTOR (both its interior AND its surface) must be an EQUIPOTENTIAL -- that is, V is CONSTANT everywhere throughout the conductor?

- **A)** This proof requires INTEGRATING (rather than differentiating) the zero-field condition, contradicting the correct direct consequence of E=-dV/dr with E=0
- **B)** A conductor's potential VARIES significantly from point to point even in electrostatic equilibrium, contradicting the actual equipotential nature of a conductor implied by its zero interior field
- **C)** This concept has no actual relationship between a conductor's zero interior field and whether it is an equipotential
- **D)** Since E=0 EVERYWHERE inside the conductor, and E=-dV/dr, it follows that dV/dr=0 throughout the conductor's interior -- meaning V does NOT change from point to point, so the ENTIRE conductor (interior and surface) must be at the SAME constant potential, a direct calculus consequence of the zero-field result

What is CAPACITANCE, C=Q/V (the ratio of stored charge to potential difference), and why is C a CONSTANT for a GIVEN capacitor GEOMETRY -- that is, why does DOUBLING the charge Q ALSO double V, leaving the RATIO Q/V unchanged?

- **A)** This proportionality argument requires the capacitor to have a SPECIFIC shape (e.g., only parallel plates), contradicting the actual GENERAL validity of C=Q/V being geometry-dependent-but-charge-independent for ANY capacitor shape
- **B)** Since the electric field (and therefore the potential difference V, found by integrating E) is DIRECTLY PROPORTIONAL to the charge Q creating it (by superposition/linearity of Coulomb's law), DOUBLING Q also DOUBLES V, keeping the ratio C=Q/V CONSTANT -- capacitance therefore depends ONLY on the capacitor's GEOMETRY (shape, size, and separation of its conductors), not on the specific amount of charge stored
- **C)** This concept has no actual relationship between the linearity of Coulomb's law and why capacitance remains constant regardless of the stored charge
- **D)** Capacitance CHANGES depending on how much charge is stored on the capacitor, contradicting the actual constant, geometry-dependent nature of C

How is the CAPACITANCE of a PARALLEL-PLATE CAPACITOR, C=epsilon_0*A/d (plate area A, separation d), DERIVED by COMBINING the infinite-plane field result, E=sigma/epsilon_0 (TWO plates DOUBLE this to E=sigma/epsilon_0 between them), with V=E*d (from integrating E over the uniform field between the plates)?

- **A)** This derivation shows capacitance INCREASING with SEPARATION d (rather than decreasing), contradicting the actual inverse relationship between C and d
- **B)** The parallel-plate capacitance formula has no actual relationship to the infinite-plane electric field result derived in the earlier Electrostatics unit
- **C)** Between the two oppositely charged plates, the fields from EACH plate ADD (by superposition) to give E=sigma/epsilon_0 (using sigma=Q/A); integrating this CONSTANT field over the separation d gives V=E*d=Qd/(epsilon_0*A); solving C=Q/V then gives C=epsilon_0*A/d -- directly derived by combining the earlier infinite-plane field result with a simple integration for V
- **D)** This derivation produces C=epsilon_0*d/A (inverting the correct area-to-separation ratio), contradicting the correct formula, C=epsilon_0*A/d

How is the CAPACITANCE of a CYLINDRICAL CAPACITOR (two coaxial conducting cylinders of radii a

How is the CAPACITANCE of a SPHERICAL CAPACITOR (two concentric conducting spheres of radii a

How is the ENERGY STORED in a CHARGED CAPACITOR, U=(1/2)*C*V^2, DERIVED using an INTEGRAL, W=∫[0 to Q] V(q) dq = ∫[0 to Q] (q/C) dq, representing the CUMULATIVE work needed to INCREMENTALLY move charge onto the capacitor's plates?

- **A)** As charge builds up on the capacitor's plates, the voltage at any intermediate charge q is V(q)=q/C (from the capacitance definition); the TOTAL work to charge the capacitor from 0 to Q is W=∫[0 to Q] (q/C) dq = Q^2/(2C), which (substituting Q=CV) can be rewritten as U=(1/2)*C*V^2 -- this integral accounts for the fact that the voltage (and therefore the work needed per unit charge) INCREASES as the capacitor charges up, rather than remaining constant
- **B)** This integration produces U=C*V^2 (missing the factor of 1/2), contradicting the correct energy formula, U=(1/2)*C*V^2
- **C)** This derivation incorrectly assumes voltage remains CONSTANT throughout the charging process, contradicting the actual increasing voltage (V=q/C) that necessitates the integral in the first place
- **D)** The energy stored in a capacitor has no actual relationship to integrating the work done while incrementally charging it

How does the ENERGY-DENSITY formula from the Electrostatics unit, u=(1/2)*epsilon_0*E^2, CONSISTENTLY REPRODUCE the parallel-plate capacitor's total stored energy, U=(1/2)*C*V^2, when the energy density is MULTIPLIED by the VOLUME between the plates (Volume=A*d)?

- **A)** This derivation requires using the energy density formula from the Gravitation unit (rather than the Electrostatics unit), contradicting the actual electrostatics-based origin of the energy density concept
- **B)** This substitution produces a result that does NOT match the capacitor energy formula, U=(1/2)CV^2, contradicting the actual consistency between these two energy-calculation approaches
- **C)** The field-energy-density formula has no actual relationship to the parallel-plate capacitor's total stored energy
- **D)** Substituting E=V/d (from V=Ed) into u=(1/2)*epsilon_0*E^2 gives u=(1/2)*epsilon_0*(V/d)^2; multiplying by the volume between the plates, Volume=A*d, gives U=u*Volume=(1/2)*epsilon_0*(V^2/d^2)*A*d=(1/2)*(epsilon_0*A/d)*V^2=(1/2)*C*V^2 -- EXACTLY matching the capacitor energy formula, confirming that the field-energy-density concept and the capacitor-energy formula are FULLY CONSISTENT