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  2. MCAT Chemical and Physical Foundations of Biological Systems
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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.

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What this deck covers

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.

How to use these flashcards

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.

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

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QUESTION

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

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ANSWER

E⃗=F⃗q\vec{E} = \frac{\vec{F}}{q}E=qF​. The electric field is defined as the electrostatic force per unit positive test charge at a point.

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Flashcard 1: State the definition of electric field in terms of force on a test charge.

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

Flashcard 2: Which direction does E⃗\vec{E}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 3: State the definition of electric potential difference in terms of potential energy change and charge.

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

Flashcard 4: State the relationship between work done by the electric field and potential difference for charge qqq.

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

Flashcard 5: Identify the relationship between electric field magnitude and potential gradient in 1D along xxx.

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

Flashcard 6: State the magnitude of the uniform electric field between parallel plates with potential difference ΔV\Delta VΔV and spacing ddd.

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

Flashcard 7: 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 8: Identify the rule relating electric field lines to equipotential surfaces.

Answer: Field lines intersect equipotentials at 90∘90^\circ90∘. The field direction is perpendicular to surfaces of constant potential, following the gradient.

Flashcard 9: State the electric potential (voltage) due to a point charge QQQ at distance rrr (zero at infinity).

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

Flashcard 10: State the value of the elementary charge magnitude, eee, in coulombs.

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

Flashcard 11: 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 12: What is the relationship between kkk, ε0\varepsilon_0ε0​, and π\piπ?

Answer: k=14πε0k = \frac{1}{4\pi\varepsilon_0}k=4πε0​1​. Coulomb's constant relates to the vacuum permittivity through this expression in electrostatics.

Flashcard 13: State Coulomb's law for the magnitude of the electrostatic force between two point charges.

Answer: F=k∣q1q2∣r2F = k\frac{|q_1 q_2|}{r^2}F=kr2∣q1​q2​∣​. The law states that the force is proportional to the product of charges and inversely proportional to the square of the distance.

Flashcard 14: What is the SI unit of electric field magnitude EEE?

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

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

Answer: k≈9.0×109 N\cdotpm2/C2k \approx 9.0 \times 10^9\ \text{N·m}^2/\text{C}^2k≈9.0×109 N\cdotpm2/C2. This value is derived from fundamental constants and used in Coulomb's law for vacuum.

Flashcard 16: How does the electrostatic force magnitude change if the separation doubles: r→2rr \to 2rr→2r?

Answer: F→F4F \to \frac{F}{4}F→4F​. Coulomb's law's inverse square dependence on distance quarters the force when distance doubles.

Flashcard 17: Find the net electric field at the midpoint between equal charges +Q+Q+Q and +Q+Q+Q separated by 2d2d2d.

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

Flashcard 18: Find the net electric field at the midpoint between charges +Q+Q+Q and −Q-Q−Q separated by 2d2d2d (direction only).

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

Flashcard 19: Identify the sign of ΔU\Delta UΔU when a positive charge moves in the direction of E⃗\vec{E}E.

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

Flashcard 20: Which direction does E⃗\vec{E}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 21: State the formula for the electric field magnitude due to a point charge QQQ a distance rrr away.

Answer: E=k∣Q∣r2E = k\frac{|Q|}{r^2}E=kr2∣Q∣​. The expression comes from dividing the Coulomb force on a test charge by its magnitude.

Flashcard 22: State the superposition principle for the net electric field from multiple charges.

Answer: E⃗net=∑iE⃗i\vec{E}_{\text{net}} = \sum_i \vec{E}_iEnet​=∑i​Ei​. Electric fields obey vector addition, allowing the total field to be the sum of individual contributions.

Flashcard 23: State the formula for electric potential energy of two point charges separated by rrr (zero at infinity).

Answer: U=kq1q2rU = k\frac{q_1 q_2}{r}U=krq1​q2​​. This represents the work to assemble the charges from infinite separation, positive for like charges.

Flashcard 24: What is the SI unit of electric potential VVV?

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

Flashcard 25: How does the electric field magnitude from a point charge change if distance triples: r→3rr \to 3rr→3r?

Answer: E→E9E \to \frac{E}{9}E→9E​. The inverse square law reduces the field to one-ninth when distance increases by a factor of three.