MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4c Electric Potential Voltage Capacitance

Study 4c Electric Potential Voltage Capacitance 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 Electric Potential Voltage Capacitance

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QUESTION
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How does inserting a dielectric with constant κ\kappa change parallel-plate capacitance?

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ANSWER

C=κε0AdC=\kappa\frac{\varepsilon_0A}{d}. Dielectric insertion reduces the effective field, increasing capacitance by the factor κ\kappa.

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Flashcard 1: How does inserting a dielectric with constant κ\kappa change parallel-plate capacitance?

Answer: C=κε0AdC=\kappa\frac{\varepsilon_0A}{d}. Dielectric insertion reduces the effective field, increasing capacitance by the factor κ\kappa.

Flashcard 2: What is the SI unit of capacitance?

Answer: 1 F=1 C/V1\ \text{F}=1\ \text{C/V}. The farad measures capacitance as one coulomb of charge stored per volt of potential difference.

Flashcard 3: State the relation between electric field and electric potential in one dimension.

Answer: E=dVdxE=-\frac{dV}{dx}. The electric field is the negative rate of change of potential with respect to position.

Flashcard 4: If V(r)=kQrV(r)=\frac{kQ}{r}, what is the ratio V(2r)V(r)\frac{V(2r)}{V(r)}?

Answer: V(2r)V(r)=12\frac{V(2r)}{V(r)}=\frac{1}{2}. Potential inversely proportional to distance, so doubling rr halves the potential.

Flashcard 5: Find QQ stored on a capacitor with C=5 μFC=5\ \mu\text{F} and V=12 VV=12\ \text{V}.

Answer: Q=CV=60 μCQ=CV=60\ \mu\text{C}. Charge stored is the product of capacitance and applied voltage.

Flashcard 6: Identify CeqC_{\text{eq}} for two identical capacitors CC in series.

Answer: Ceq=C2C_{\text{eq}}=\frac{C}{2}. For identical capacitors in series, equivalent is half due to doubled effective separation.

Flashcard 7: State the formula for potential difference between two points in terms of work.

Answer: ΔV=Wfieldq\Delta V=-\frac{W_{\text{field}}}{q}. Potential difference equals the negative work done by the field per unit charge when moving a charge between points.

Flashcard 8: Which has zero work by the electric field: motion along or perpendicular to an equipotential surface?

Answer: Perpendicular motion has W=0W=0 (along equipotential). Motion perpendicular to the field lines follows equipotential surfaces, resulting in no change in potential and zero work.

Flashcard 9: What is the electric potential energy of two point charges q1q_1 and q2q_2 separated by rr?

Answer: U=kq1q2rU=\frac{kq_1q_2}{r}. Represents the work to assemble the charges from infinity, analogous to gravitational potential energy.

Flashcard 10: What is the definition of electric potential VV at a point?

Answer: V=UqV=\frac{U}{q} (electric potential energy per unit charge). Electric potential at a point is the electric potential energy per unit charge for a test charge placed there.

Flashcard 11: What is the SI unit of electric potential (voltage)?

Answer: 1 V=1 J/C1\ \text{V}=1\ \text{J/C}. The volt is defined as the potential difference that imparts one joule of energy to one coulomb of charge.

Flashcard 12: What is the definition of capacitance CC in terms of charge and voltage?

Answer: C=QΔVC=\frac{Q}{\Delta V}. Capacitance quantifies the ability to store charge for a given potential difference across the device.

Flashcard 13: State the energy stored in a capacitor in terms of QQ and CC.

Answer: U=Q22CU=\frac{Q^2}{2C}. Alternative form obtained by substituting Q=CVQ=CV into the energy expression.

Flashcard 14: What is the superposition rule for electric potential from multiple charges?

Answer: Vnet=ikQiriV_{\text{net}}=\sum_i \frac{kQ_i}{r_i}. Electric potential is a scalar quantity, allowing direct summation of individual contributions.

Flashcard 15: State the equivalent capacitance for capacitors in series.

Answer: 1Ceq=i1Ci\frac{1}{C_{\text{eq}}}=\sum_i \frac{1}{C_i}. In series, total voltage divides across capacitors, leading to reciprocal sum for equivalent capacitance.

Flashcard 16: State the equivalent capacitance for capacitors in parallel.

Answer: Ceq=iCiC_{\text{eq}}=\sum_i C_i. In parallel, charges add while sharing the same voltage, summing individual capacitances.

Flashcard 17: Identify CeqC_{\text{eq}} for two identical capacitors CC in parallel.

Answer: Ceq=2CC_{\text{eq}}=2C. Parallel connection doubles effective plate area for identical capacitors, doubling capacitance.

Flashcard 18: State the relationship between change in electric potential energy and voltage.

Answer: ΔU=qΔV\Delta U=q\Delta V. Change in potential energy equals charge times the potential difference experienced by the charge.

Flashcard 19: Which quantity is continuous across a conductor's surface in electrostatic equilibrium: VV or EE?

Answer: VV is constant throughout the conductor. In electrostatic equilibrium, the potential is uniform inside and on the surface of a conductor due to zero internal field.

Flashcard 20: What is the electric potential due to a point charge QQ at distance rr?

Answer: V=kQrV=\frac{kQ}{r}. Derived from integrating the electric field from infinity to rr, assuming zero potential at infinity.

Flashcard 21: State the energy stored in a capacitor in terms of CC and VV.

Answer: U=12CV2U=\frac{1}{2}CV^2. Energy stored derives from the work to charge the capacitor, integrating QdVQ dV from 0 to VV.

Flashcard 22: In a uniform electric field, what is the relation between ΔV\Delta V, EE, and displacement Δx\Delta x along the field?

Answer: ΔV=EΔx\Delta V=-E\Delta x. In a uniform field, potential decreases linearly in the direction of the field by the product of field strength and distance.

Flashcard 23: State the capacitance of a parallel-plate capacitor with plate area AA and separation dd (vacuum).

Answer: C=ε0AdC=\frac{\varepsilon_0A}{d}. For parallel plates, capacitance increases with area and decreases with separation, proportional to permittivity.

Flashcard 24: Find the energy stored when C=2 μFC=2\ \mu\text{F} and V=3 VV=3\ \text{V}.

Answer: U=12CV2=9 μJU=\frac{1}{2}CV^2=9\ \mu\text{J}. Energy calculation uses the formula derived from integrating work done during charging.

Flashcard 25: What is the sign of VV at a point due to a negative source charge Q<0Q<0?

Answer: V<0V<0. For Q<0Q<0, potential is negative relative to zero at infinity, indicating attractive interaction for positive test charges.