MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4c Electrostatics Electric Fields

Study 4c Electrostatics Electric Fields in MCAT Chemical and Physical Foundations of Biological Systems with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

MCAT Chemical and Physical Foundations of Biological Systems

4c Electrostatics Electric Fields

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What is the SI unit of electric potential VV?

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ANSWER

Volt (V) = J/C\text{J/C}. The volt equals one joule of energy per coulomb of charge.

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This deck focuses on 4c Electrostatics Electric Fields, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Flashcard 1: What is the SI unit of electric potential VV?

Answer: Volt (V) = J/C\text{J/C}. The volt equals one joule of energy per coulomb of charge.

Flashcard 2: State the superposition principle for the net electric field from multiple charges.

Answer: Enet=iEi\vec{E}_{\text{net}} = \sum_i \vec{E}_i. Electric fields obey vector addition, allowing the total field to be the sum of individual contributions.

Flashcard 3: Find the net electric field at the midpoint between charges +Q+Q and Q-Q separated by 2d2d (direction only).

Answer: Toward the Q-Q charge (from +Q+Q to Q-Q). Fields from opposite charges at the midpoint add constructively, pointing towards the negative charge.

Flashcard 4: State the formula for the electric field magnitude due to a point charge QQ a distance rr away.

Answer: E=kQr2E = k\frac{|Q|}{r^2}. The expression comes from dividing the Coulomb force on a test charge by its magnitude.

Flashcard 5: State the relationship between work done by the electric field and potential difference for charge qq.

Answer: Wfield=qΔVW_{\text{field}} = -q\Delta V. For conservative fields, work by the field equals the negative change in potential energy.

Flashcard 6: Identify the relationship between electric field magnitude and potential gradient in 1D along xx.

Answer: Ex=dVdxE_x = -\frac{dV}{dx}. The electric field component is the negative rate of change of potential with position.

Flashcard 7: State the value of the elementary charge magnitude, ee, in coulombs.

Answer: e=1.60×1019 Ce = 1.60 \times 10^{-19}\ \text{C}. This magnitude represents the fundamental unit of charge carried by a proton or electron.

Flashcard 8: Identify the sign of ΔU\Delta U when a positive charge moves in the direction of E\vec{E}.

Answer: ΔU<0\Delta U < 0. Moving along the field decreases potential for positive charges, reducing potential energy.

Flashcard 9: State the definition of electric field in terms of force on a test charge.

Answer: E=Fq\vec{E} = \frac{\vec{F}}{q}. The electric field is defined as the electrostatic force per unit positive test charge at a point.

Flashcard 10: State the formula for electric potential energy of two point charges separated by rr (zero at infinity).

Answer: U=kq1q2rU = k\frac{q_1 q_2}{r}. This represents the work to assemble the charges from infinite separation, positive for like charges.

Flashcard 11: What is the relationship between kk, ε0\varepsilon_0, and π\pi?

Answer: k=14πε0k = \frac{1}{4\pi\varepsilon_0}. Coulomb's constant relates to the vacuum permittivity through this expression in electrostatics.

Flashcard 12: State the definition of electric potential difference in terms of potential energy change and charge.

Answer: ΔV=ΔUq\Delta V = \frac{\Delta U}{q}. Potential difference measures the change in potential energy per unit charge moved between points.

Flashcard 13: State the electric potential (voltage) due to a point charge QQ at distance rr (zero at infinity).

Answer: V=kQrV = k\frac{Q}{r}. The potential is the work per unit charge to bring a test charge from infinity to that point.

Flashcard 14: Which direction does E\vec{E} point around an isolated positive point charge?

Answer: Radially outward from the charge. The field direction indicates the force on a positive test charge, repelling from a positive source.

Flashcard 15: What is the approximate value of Coulomb's constant, kk, in vacuum?

Answer: k9.0×109 N\cdotpm2/C2k \approx 9.0 \times 10^9\ \text{N·m}^2/\text{C}^2. This value is derived from fundamental constants and used in Coulomb's law for vacuum.

Flashcard 16: Which direction does E\vec{E} point around an isolated negative point charge?

Answer: Radially inward toward the charge. The field points in the direction a positive test charge would be forced, attracting to a negative source.

Flashcard 17: State the magnitude of the uniform electric field between parallel plates with potential difference ΔV\Delta V and spacing dd.

Answer: E=ΔVdE = \frac{\Delta V}{d}. In uniform fields, the field strength equals the potential gradient across the plate separation.

Flashcard 18: Identify the rule relating electric field lines to equipotential surfaces.

Answer: Field lines intersect equipotentials at 9090^\circ. The field direction is perpendicular to surfaces of constant potential, following the gradient.

Flashcard 19: State Coulomb's law for the magnitude of the electrostatic force between two point charges.

Answer: F=kq1q2r2F = k\frac{|q_1 q_2|}{r^2}. The law states that the force is proportional to the product of charges and inversely proportional to the square of the distance.

Flashcard 20: How does the electrostatic force magnitude change if the separation doubles: r2rr \to 2r?

Answer: FF4F \to \frac{F}{4}. Coulomb's law's inverse square dependence on distance quarters the force when distance doubles.

Flashcard 21: What is the direction of the electric field between parallel plates: from ++ plate to - plate or the reverse?

Answer: From the ++ plate toward the - plate. The field points from higher to lower potential, consistent with force on positive charges.

Flashcard 22: Find the net electric field at the midpoint between equal charges +Q+Q and +Q+Q separated by 2d2d.

Answer: Enet=0E_{\text{net}} = 0. At the midpoint, fields from identical charges are equal in magnitude but opposite in direction, canceling out.

Flashcard 23: How does the electric field magnitude from a point charge change if distance triples: r3rr \to 3r?

Answer: EE9E \to \frac{E}{9}. The inverse square law reduces the field to one-ninth when distance increases by a factor of three.

Flashcard 24: What is the SI unit of electric charge?

Answer: Coulomb (C). The coulomb quantifies electric charge in the SI system as the charge transported by a constant current of one ampere in one second.

Flashcard 25: What is the SI unit of electric field magnitude EE?

Answer: N/C\text{N/C} (equivalently V/m\text{V/m}). The unit derives from force per unit charge or equivalently potential difference per unit distance.