AP Physics 2 Flashcards: Electric Charge And Electric Force

Study Electric Charge And Electric Force in AP Physics 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Physics 2

Electric Charge And Electric Force

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QUESTION
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What is the direction of the electric field due to a positive charge?

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ANSWER

Radially outward from the charge. Field lines point away from positive charges.

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This deck focuses on Electric Charge And Electric Force, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.

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Flashcard 1: What is the direction of the electric field due to a positive charge?

Answer: Radially outward from the charge. Field lines point away from positive charges.

Flashcard 2: Find the force on a charge in a uniform electric field.

Answer: F=qEF = qE. Force equals charge times field strength.

Flashcard 3: Identify the SI unit for electric field strength.

Answer: Newton per Coulomb (N/C). Also equivalent to V/m (volts per meter).

Flashcard 4: What is the electric field on the surface of a charged conductor?

Answer: Perpendicular to the surface. Field is normal to equipotential surfaces.

Flashcard 5: What is the electric field on the surface of a charged conductor?

Answer: Perpendicular to the surface. Field is normal to equipotential surfaces.

Flashcard 6: What is the direction of the electric field due to a negative charge?

Answer: Radially inward toward the charge. Field lines point toward negative charges.

Flashcard 7: State the formula for electric force between two charges.

Answer: F=kq1q2r2F = k \frac{|q_1 q_2|}{r^2}. Coulomb's Law for electrostatic force.

Flashcard 8: What is the effect of a charged insulator on a nearby uncharged conductor?

Answer: Induces a charge separation. Creates surface charges that cancel internal field.

Flashcard 9: Find the force on a charge in a uniform electric field.

Answer: F=qEF = qE. Force equals charge times field strength.

Flashcard 10: What is the charge of an electron?

Answer: 1.6×1019 C-1.6 \times 10^{-19} \text{ C}. Fundamental negative charge magnitude.

Flashcard 11: What is the principle of superposition for electric forces?

Answer: Net force is vector sum of individual forces. Forces add vectorially from all charges.

Flashcard 12: What is the electric potential at a point due to a point charge?

Answer: V=kqrV = k \frac{q}{r}. Potential decreases with distance from positive charge.

Flashcard 13: State the relationship between potential energy and work done by electric forces.

Answer: UiUf=WdoneU_i - U_f = W_{\text{done}}. Conservation of energy in electric fields.

Flashcard 14: What is the formula for the potential energy of a dipole in an electric field?

Answer: U=pEcos(θ)U = -pE \text{cos}(\theta). Energy depends on dipole orientation relative to field.

Flashcard 15: State the formula for electric force between two charges.

Answer: F=kq1q2r2F = k \frac{|q_1 q_2|}{r^2}. Coulomb's Law for electrostatic force.

Flashcard 16: Determine the potential difference between two points in an electric field.

Answer: V=Ed\triangle V = Ed. For uniform field, voltage change is field times distance.

Flashcard 17: Calculate the electric potential at 2 m2 \text{ m} from a 3 C3 \text{ C} charge.

Answer: V=1.35×1010 VV = 1.35 \times 10^{10} \text{ V}. Using V=kqrV = k\frac{q}{r} with given values.

Flashcard 18: Calculate the force between two 1 C1 \text{ C} charges 1 m1 \text{ m} apart.

Answer: F=8.99×109 NF = 8.99 \times 10^9 \text{ N}. Using F=kq1q2r2F = k\frac{|q_1 q_2|}{r^2} with given values.

Flashcard 19: What is the electric potential at a point due to a point charge?

Answer: V=kqrV = k \frac{q}{r}. Potential decreases with distance from positive charge.

Flashcard 20: Determine the electric force on a 5 C5 \text{ C} charge in a 10 N/C10 \text{ N/C} field.

Answer: F=50 NF = 50 \text{ N}. Using F=qEF = qE with given charge and field.

Flashcard 21: State Coulomb's Law.

Answer: F=kq1q2r2F = k \frac{|q_1 q_2|}{r^2}. Force between charges is proportional to product of charges, inversely proportional to distance squared.

Flashcard 22: Determine the electric force on a 5 C5 \text{ C} charge in a 10 N/C10 \text{ N/C} field.

Answer: F=50 NF = 50 \text{ N}. Using F=qEF = qE with given charge and field.

Flashcard 23: What is the electric field inside a conductor in electrostatic equilibrium?

Answer: Zero. Charges redistribute to cancel internal field.

Flashcard 24: What is the net charge inside a conductor in electrostatic equilibrium?

Answer: Zero inside the conductor. All excess charge resides on surface.

Flashcard 25: Identify the relationship between electric field and force on a charged particle.

Answer: F=qEF = qE. Definition of electric field strength.

Flashcard 26: State Gauss's Law.

Answer: Flux=Qencε0\text{Flux} = \frac{Q_{\text{enc}}}{\varepsilon_0}. Electric flux through closed surface equals enclosed charge over ε0\varepsilon_0.

Flashcard 27: Calculate the work done moving a 1 C1 \text{ C} charge across 10 V10 \text{ V}.

Answer: W=10 JW = 10 \text{ J}. Work equals charge times potential difference.

Flashcard 28: Calculate the electric potential at 2 m2 \text{ m} from a 3 C3 \text{ C} charge.

Answer: V=1.35×1010 VV = 1.35 \times 10^{10} \text{ V}. Using V=kqrV = k\frac{q}{r} with given values.

Flashcard 29: What is meant by a conservative electric field?

Answer: Work done is path independent. Electric fields are conservative force fields.

Flashcard 30: What is the principle of superposition for electric forces?

Answer: Net force is vector sum of individual forces. Forces add vectorially from all charges.

Flashcard 31: What is the relationship between electric field and voltage?

Answer: E=dVdxE = -\frac{dV}{dx}. Field points from high to low potential.

Flashcard 32: Calculate the force between two 1 C1 \text{ C} charges 1 m1 \text{ m} apart.

Answer: F=8.99×109 NF = 8.99 \times 10^9 \text{ N}. Using F=kq1q2r2F = k\frac{|q_1 q_2|}{r^2} with given values.

Flashcard 33: What is the magnitude of the electric field in a parallel plate capacitor?

Answer: E=VdE = \frac{\text{V}}{d}. Uniform field between parallel plates.

Flashcard 34: Calculate the work done moving a 1 C1 \text{ C} charge across 10 V10 \text{ V}.

Answer: W=10 JW = 10 \text{ J}. Work equals charge times potential difference.

Flashcard 35: What is the permittivity of free space, ε0\text{ε}_0?

Answer: 8.85×1012 C2/N m28.85 \times 10^{-12} \text{ C}^2/\text{N m}^2. Fundamental constant in electromagnetism.

Flashcard 36: Define electric field.

Answer: Force per unit charge, E=FqE = \frac{F}{q}. Electric force per unit test charge.

Flashcard 37: Identify the formula for the work done by an electric force.

Answer: W=qEdW = qEd. Work equals force times distance moved.

Flashcard 38: What are the units of electric potential (voltage)?

Answer: Volt (V). One volt equals one joule per coulomb.

Flashcard 39: What is the charge of a proton?

Answer: +1.6×1019 C6 \times 10^{-19} \text{ C}. Fundamental positive charge magnitude.

Flashcard 40: State Coulomb's Law.

Answer: F=kq1q2r2F = k \frac{|q_1 q_2|}{r^2}. Force between charges is proportional to product of charges, inversely proportional to distance squared.

Flashcard 41: What is the formula for electric potential energy?

Answer: U=kq1q2rU = k \frac{q_1 q_2}{r}. Energy increases when like charges approach.

Flashcard 42: What is the formula for the potential energy of a dipole in an electric field?

Answer: U=pEcos(θ)U = -pE \text{cos}(\theta). Energy depends on dipole orientation relative to field.

Flashcard 43: Determine the potential difference between two points in an electric field.

Answer: V=Ed\triangle V = Ed. For uniform field, voltage change is field times distance.

Flashcard 44: State the formula for electric potential (voltage).

Answer: V=UqV = \frac{U}{q}. Electric potential energy per unit charge.

Flashcard 45: Calculate the potential energy of a 2 C2 \text{ C} charge at 5 V5 \text{ V}.

Answer: U=10 JU = 10 \text{ J}. Using U=qVU = qV with given charge and potential.

Flashcard 46: What is the magnitude of the electric field in a parallel plate capacitor?

Answer: E=VdE = \frac{\text{V}}{d}. Uniform field between parallel plates.

Flashcard 47: What is the formula for electric field due to a point charge?

Answer: E=kqr2E = k \frac{|q|}{r^2}. Field strength decreases with distance squared.

Flashcard 48: What is the formula for electric field due to a point charge?

Answer: E=kqr2E = k \frac{|q|}{r^2}. Field strength decreases with distance squared.

Flashcard 49: What is the charge of an electron?

Answer: -1.6×1019 C6 \times 10^{-19} \text{ C}. Fundamental negative charge magnitude.

Flashcard 50: What is the net charge inside a conductor in electrostatic equilibrium?

Answer: Zero inside the conductor. All excess charge resides on surface.

Flashcard 51: What is the charge of a proton?

Answer: +1.6×1019 C6 \times 10^{-19} \text{ C}. Fundamental positive charge magnitude.

Flashcard 52: State the formula for the torque on a dipole in a uniform electric field.

Answer: τ=pEsin(θ)\tau = pE \text{sin}(\theta). Maximum torque when dipole perpendicular to field.

Flashcard 53: What is the value of the Coulomb constant, kk?

Answer: 8.99×109 N m2/C28.99 \times 10^9 \text{ N m}^2/\text{C}^2. Also written as 14πε0\frac{1}{4\pi\varepsilon_0}.

Flashcard 54: Identify the formula for the work done by an electric force.

Answer: W=qEdW = qEd. Work equals force times distance moved.

Flashcard 55: Define electric field.

Answer: Force per unit charge, E=FqE = \frac{F}{q}. Electric force per unit test charge.

Flashcard 56: Calculate the potential energy of a 2 C2 \text{ C} charge at 5 V5 \text{ V}.

Answer: U=10 JU = 10 \text{ J}. Using U=qVU = qV with given charge and potential.

Flashcard 57: What is the electric field inside a conductor in electrostatic equilibrium?

Answer: Zero. Charges redistribute to cancel internal field.

Flashcard 58: What is the relationship between electric field and voltage?

Answer: E=dVdxE = -\frac{dV}{dx}. Field points from high to low potential.

Flashcard 59: What is the effect of a charged insulator on a nearby uncharged conductor?

Answer: Induces a charge separation. Creates surface charges that cancel internal field.

Flashcard 60: State the formula for the torque on a dipole in a uniform electric field.

Answer: τ=pEsin(θ)\tau = pE \text{sin}(\theta). Maximum torque when dipole perpendicular to field.

Flashcard 61: State the formula for electric potential (voltage).

Answer: V=UqV = \frac{U}{q}. Electric potential energy per unit charge.

Flashcard 62: What are the units of electric potential (voltage)?

Answer: Volt (V). One volt equals one joule per coulomb.

Flashcard 63: What is the permittivity of free space, ε0\text{ε}_0?

Answer: 8.85×1012 C2/N m28.85 \times 10^{-12} \text{ C}^2/\text{N m}^2. Fundamental constant in electromagnetism.

Flashcard 64: What is the direction of the electric field due to a negative charge?

Answer: Radially inward toward the charge. Field lines point toward negative charges.

Flashcard 65: State the relationship between potential energy and work done by electric forces.

Answer: UiUf=WdoneU_i - U_f = W_{\text{done}}. Conservation of energy in electric fields.

Flashcard 66: What is the direction of the electric field due to a positive charge?

Answer: Radially outward from the charge. Field lines point away from positive charges.

Flashcard 67: Identify the SI unit for electric field strength.

Answer: Newton per Coulomb (N/C). Also equivalent to V/m (volts per meter).

Flashcard 68: Identify the relationship between electric field and force on a charged particle.

Answer: F=qEF = qE. Definition of electric field strength.

Flashcard 69: What is the value of the Coulomb constant, kk?

Answer: 8.99×109 N m2/C28.99 \times 10^9 \text{ N m}^2/\text{C}^2. Also written as 14πε0\frac{1}{4\pi\varepsilon_0}.

Flashcard 70: What is meant by a conservative electric field?

Answer: Work done is path independent. Electric fields are conservative force fields.