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This deck focuses on Electric Fields, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.
Study Electric Fields in AP Physics 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Explain the change in electric field strength if the charge is tripled.
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Field strength is tripled. Field strength is directly proportional to the source charge magnitude.
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This deck focuses on Electric Fields, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: Field strength is tripled. Field strength is directly proportional to the source charge magnitude.
Answer: E=dV. Uniform field between plates equals voltage divided by separation.
Answer: Zero. Field approaches zero as distance approaches infinity for finite charges.
Answer: Distorts and reduces the field inside. Conductor redistributes charges and creates field-free interior region.
Answer: Radially inward towards the charge. Electric field points toward negative charges in all directions.
Answer: −1.6×10−19 C. Elementary negative charge magnitude in coulombs.
Answer: Field strength is tripled. Field strength is directly proportional to the source charge magnitude.
Answer: Field is independent of the plate area. Field depends only on charge density and plate separation, not area.
Answer: Zero. Free charges redistribute to cancel internal fields in equilibrium.
Answer: Field strength decreases by a factor of 4. Inverse square relationship: doubling distance quarters the field strength.
Answer: ∮E⋅dA=ε0Qenclosed. Fundamental relationship between electric flux and enclosed charge.
Answer: Zero, if charges are opposite. Equal magnitudes but opposite directions cancel when charges are opposite.
Answer: From positive to negative charge. Field lines exit positive charges and enter negative charges.
Answer: Perpendicular to the surface. Electric field must be normal to conductor surface in equilibrium.
Answer: Zero. Field approaches zero as distance approaches infinity for finite charges.
Answer: Zero. Free charges redistribute to cancel internal fields in equilibrium.
Answer: Principle of superposition. Vector addition of individual field contributions from multiple sources.
Answer: Field lines terminate on the conductor. Grounded conductors provide discharge path, terminating field lines.
Answer: −1.6×10−19 C. Elementary negative charge magnitude in coulombs.
Answer: Radially inward towards the charge. Electric field points toward negative charges in all directions.
Answer: Distorts and reduces the field inside. Conductor redistributes charges and creates field-free interior region.
Answer: Perpendicular to the surface. Electric field must be normal to conductor surface in equilibrium.
Answer: Field lines terminate on the conductor. Grounded conductors provide discharge path, terminating field lines.
Answer: F=r2k∣q1q2∣. Force between charges depends on product of charges and inverse square of distance.
Answer: k=8.99×109 N m2/C2. Standard value for the proportionality constant in vacuum.
Answer: ∮E⋅dA=ε0Qenclosed. Fundamental relationship between electric flux and enclosed charge.
Answer: Newton per coulomb (N/C). Standard SI unit combining force (N) per unit charge (C).
Answer: Electric field decreases by a factor of 4. Inverse square law: field strength decreases by factor of four.
Answer: Radially outward from the charge. Electric field points away from positive charges in all directions.
Answer: E=dV. Uniform field between plates equals voltage divided by separation.
Answer: E=r2kQ. Coulomb's law applied to field strength, where k is the electric constant.
Answer: Zero. No charges inside hollow conductor means zero internal field.
Answer: Electric field decreases by a factor of 4. Inverse square law: field strength decreases by factor of four.
Answer: E=−dxdV. Field is the negative gradient of potential in one dimension.
Answer: Two equal and opposite charges separated by a distance. Basic configuration creating electric dipole moment and field pattern.
Answer: Field strength decreases by a factor of 4. Inverse square relationship: doubling distance quarters the field strength.
Answer: F=r2k∣q1q2∣. Force between charges depends on product of charges and inverse square of distance.
Answer: From positive to negative charge. Field lines exit positive charges and enter negative charges.
Answer: Reduces the electric field. Dielectric materials decrease field strength by polarization effects.
Answer: Zero. Symmetry causes all field contributions to cancel at the center.
Answer: Zero, if charges are opposite. Equal magnitudes but opposite directions cancel when charges are opposite.
Answer: Not necessarily zero. Field and potential are related but independent quantities.
Answer: Electric potential. Potential represents energy per unit charge in electric field.
Answer: E=−dxdV. Field is the negative gradient of potential in one dimension.
Answer: Same as a point charge: E=r2kQ. Spherical symmetry makes external field identical to point charge.
Answer: F=qE. Force equals charge multiplied by electric field strength.
Answer: Zero. Symmetry causes all field contributions to cancel at the center.
Answer: Zero. Symmetry causes equal and opposite field contributions to cancel.
Answer: Two equal and opposite charges separated by a distance. Basic configuration creating electric dipole moment and field pattern.
Answer: Zero. Symmetry causes equal and opposite field contributions to cancel.
Answer: Principle of superposition. Vector addition of individual field contributions from multiple sources.
Answer: Radially outward from the charge. Electric field points away from positive charges in all directions.
Answer: Same as a point charge: E=r2kQ. Spherical symmetry makes external field identical to point charge.
Answer: Reduces the electric field. Dielectric materials decrease field strength by polarization effects.
Answer: Not necessarily zero. Field and potential are related but independent quantities.
Answer: Field is independent of the plate area. Field depends only on charge density and plate separation, not area.
Answer: E=r2kQ. Coulomb's law applied to field strength, where k is the electric constant.
Answer: Newton per coulomb (N/C). Standard SI unit combining force (N) per unit charge (C).
Answer: k=8.99×109 N m2/C2. Standard value for the proportionality constant in vacuum.
Answer: Zero. No charges inside hollow conductor means zero internal field.
Answer: F=qE. Force equals charge multiplied by electric field strength.