AP Physics C: E&M:: Electrostatics

How does COULOMB'S LAW, F=kq1q2/r^2, and the resulting ELECTRIC FIELD, E=kq/r^2 (force per unit charge), MATHEMATICALLY PARALLEL Newton's law of gravitation, F=GMm/r^2, and gravitational field, g=GM/r^2, from the earlier AP Physics C: Mechanics course?

- **A)** Coulomb's law and Newton's law of gravitation have COMPLETELY UNRELATED mathematical forms, contradicting their actual shared inverse-square structure
- **B)** This parallel structure means electric field and gravitational field are measured in the SAME physical units, contradicting their actual different units (N/C versus m/s^2 or N/kg)
- **C)** Both laws share the IDENTICAL inverse-square mathematical structure (force proportional to 1/r^2), differing only in that gravity is ALWAYS attractive (mass is always positive) while the electric force can be EITHER attractive or repulsive depending on the signs of the two charges -- this parallel means many calculus techniques from Gravitation (field-vs-force distinction, potential energy via integration, shell-theorem-style symmetry arguments) carry over directly to electrostatics
- **D)** The electric force, unlike gravity, is ALWAYS attractive regardless of the signs of the charges involved, contradicting the actual charge-sign-dependent nature of the electric force

How is the ELECTRIC FIELD from a CONTINUOUS LINE of charge (a thin rod of length L, total charge Q, linear charge density lambda=Q/L) at a point ALONG its perpendicular bisector DERIVED using an INTEGRAL, E=∫dE, summing the field contributions from every infinitesimal charge element dq=lambda*dx?

- **A)** This integral produces a field EQUAL to the simple point-charge formula kq/r^2 at ALL points, contradicting the actual distance-and-geometry-dependent result of proper integration over the rod's extent
- **B)** This integral has no actual relationship to finding the electric field from a continuous, extended charge distribution like a charged rod
- **C)** Treating the rod as a continuum of infinitesimal charge elements dq=lambda*dx, each contributing a field dE=k*dq/r^2 at the point of interest (with only the component PERPENDICULAR to the rod surviving after integration, by symmetry -- the parallel components cancel in pairs), and INTEGRATING over the rod's full length gives the total field -- generalizing the point-charge formula, E=kq/r^2, to an extended, continuous charge distribution via integral calculus
- **D)** This integration technique requires treating the rod as a SINGLE point charge located at its center, contradicting the actual need for genuine integration when computing the field at a general point

How is the ELECTRIC FIELD on the AXIS of a UNIFORMLY CHARGED RING (radius R, total charge Q) DERIVED using an integral over infinitesimal charge elements dq, exploiting SYMMETRY to show that only the AXIAL component of the field SURVIVES the integration?

- **A)** The electric field on a charged ring's axis has no actual relationship to symmetry arguments or to integrating over infinitesimal charge elements
- **B)** This integral produces a field that is INDEPENDENT of the distance x along the axis, contradicting the actual x-dependent result, E_axial=kQx/(x^2+R^2)^(3/2)
- **C)** By symmetry, for every charge element dq on the ring, there is an OPPOSITE element whose PERPENDICULAR (radial) field component EXACTLY CANCELS, leaving only the AXIAL component to survive; integrating this axial component, dE_axial=k*dq*x/(x^2+R^2)^(3/2), over the entire ring gives E_axial=kQx/(x^2+R^2)^(3/2), where x is the distance along the axis from the ring's center
- **D)** This derivation shows that BOTH the axial AND radial field components survive the integration equally, contradicting the actual radial-cancellation-by-symmetry result

How is the ELECTRIC FIELD on the AXIS of a UNIFORMLY CHARGED DISK (radius R, surface charge density sigma) DERIVED by INTEGRATING the charged-ring result over many CONCENTRIC RINGS of radius r' (from r'=0 to r'=R), each treated as an infinitesimal charged ring?

- **A)** Treating the disk as a continuum of concentric rings, each of radius r' and infinitesimal charge dq=sigma*(2*pi*r')*dr', and applying the charged-ring axial-field formula to EACH ring, then INTEGRATING over r' from 0 to R gives the total axial field of the disk -- directly reusing (rather than re-deriving from scratch) the charged-ring result as the building block for a MORE complex, two-dimensional charge distribution
- **B)** This derivation requires treating the disk as a SINGLE point charge at its center, contradicting the actual need for genuine integration over concentric rings
- **C)** The charged-disk field has no actual relationship to integrating the charged-ring result over concentric rings
- **D)** This integration technique applies only to charge distributions with LINEAR (not surface) charge density, contradicting its actual applicability to a two-dimensional charged disk with surface charge density sigma

How is ELECTRIC FLUX, Phi_E=∫E·dA (a SURFACE INTEGRAL of the electric field over a closed surface), DEFINED, and what does this integral PHYSICALLY REPRESENT in terms of the number of electric field lines passing through the surface?

- **A)** Electric flux, Phi_E=∫E·dA, is the SURFACE INTEGRAL of the electric field's component PERPENDICULAR to each infinitesimal area element dA, summed (integrated) over the ENTIRE surface -- physically representing the TOTAL 'flow' of electric field lines through that surface, analogous to how a flux integral in fluid dynamics represents the flow rate of a fluid through a surface
- **B)** Electric flux is defined using a LINE integral (rather than a SURFACE integral), contradicting the actual surface-integral-based definition, Phi_E=∫E·dA
- **C)** Electric flux has no actual relationship to integrating the electric field over a surface or to the concept of field lines passing through that surface
- **D)** Electric flux depends ONLY on the surface's total AREA, regardless of the electric field's strength or direction relative to that surface, contradicting the actual field-dependent, direction-sensitive nature of the flux integral

What is GAUSS'S LAW, ∮E·dA=Q_enclosed/epsilon_0 (electric flux through any CLOSED surface equals the ENCLOSED charge divided by the permittivity of free space), and how does it provide a POWERFUL SHORTCUT for finding electric fields in HIGHLY SYMMETRIC situations, AVOIDING the need for direct integration over charge elements?

- **A)** For charge distributions with SUFFICIENT SYMMETRY (spherical, cylindrical, or planar), a well-chosen 'Gaussian surface' allows E to be pulled OUTSIDE the flux integral (since E is CONSTANT in magnitude and PARALLEL to dA everywhere on the surface), reducing ∮E·dA to simply E times the surface's area -- turning Gauss's law into an ALGEBRAIC equation for E, completely AVOIDING the need for the direct, often difficult charge-element integration used earlier in this unit
- **B)** Gauss's law can ONLY be used to find the TOTAL enclosed charge, NEVER to find the electric field itself, contradicting its actual widespread use as a field-finding shortcut for symmetric situations
- **C)** Gauss's law has no actual relationship to symmetry or to simplifying the calculation of electric fields for symmetric charge distributions
- **D)** Gauss's law requires performing a MORE DIFFICULT integration than the direct charge-element method, contradicting its actual role as a SIMPLIFYING shortcut for symmetric charge distributions

How is Gauss's law used to DERIVE the electric field OUTSIDE a UNIFORMLY CHARGED SPHERE (total charge Q, radius R) at distance r>R, by choosing a SPHERICAL Gaussian surface of radius r CONCENTRIC with the charged sphere?

- **A)** Gauss's law cannot actually be used to find the electric field outside a uniformly charged sphere, contradicting its well-established use for this exact symmetric scenario
- **B)** This derivation produces E=kQ/r (missing a power of r), contradicting the correct inverse-square result, E=kQ/r^2
- **C)** This derivation shows the electric field outside a uniformly charged sphere DEPENDS on the exact distribution of charge WITHIN the sphere (not just the total charge Q), contradicting the actual point-charge-equivalent result from Gauss's law
- **D)** By symmetry, E is CONSTANT in magnitude and points RADIALLY outward everywhere on the spherical Gaussian surface, so ∮E·dA=E*(4*pi*r^2)=Q/epsilon_0, giving E=Q/(4*pi*epsilon_0*r^2)=kQ/r^2 -- EXACTLY the same result as treating the entire charge as if concentrated at the sphere's center, directly analogous to the shell theorem's result for gravity

How does Gauss's law show that the ELECTRIC FIELD INSIDE a UNIFORMLY CHARGED SPHERICAL SHELL (a hollow shell, at any point STRICTLY inside the cavity) is EXACTLY ZERO, using the SAME reasoning as the gravitational shell theorem from the earlier Mechanics course?

- **A)** The electric field inside a charged spherical shell is EQUAL in magnitude to the field just outside the shell's surface, contradicting the actual zero-field result for the interior
- **B)** This derivation requires the shell to carry a NEGATIVE total charge for the interior field to be zero, contradicting the actual charge-sign-independent nature of this zero-interior-field result
- **C)** Gauss's law has no actual relationship to determining the electric field inside a hollow charged spherical shell
- **D)** For a spherical Gaussian surface drawn INSIDE the hollow shell (radius r < shell's radius), the ENCLOSED charge is ZERO (all the shell's charge lies OUTSIDE this Gaussian surface); since Q_enclosed=0, Gauss's law gives ∮E·dA=0, and by symmetry this forces E=0 EVERYWHERE inside the shell -- directly paralleling the zero-interior-field result for a gravitational shell