Electrostatics and Electric Fields (4C)
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MCAT Chemical and Physical Foundations of Biological Systems › Electrostatics and Electric Fields (4C)
A charged particle moves through a region with a uniform electric field but no magnetic field. The field is constant in time. Based on the setup, which outcome is most likely for the particle’s kinetic energy as it moves in the direction of the electric force (ignore collisions)?
It decreases, because electric fields always oppose motion.
It remains constant, because only magnetic fields can change speed.
It becomes zero, because uniform fields cancel the particle’s charge.
It increases, because the electric field does positive work on the particle.
Explanation
This question assesses understanding of electrostatics and electric fields in the context of kinetic energy changes for a charged particle. Electrostatics involves the study of forces between charges, described by Coulomb's law and the concept of electric fields. In this setup, the electric field is uniform, doing work on the particle along the force direction. Choice A is correct because motion along the force increases kinetic energy via positive work. Choice B is incorrect because it assumes opposition, not always true. When analyzing electric fields, always consider work W = F·d; ensure reasoning aligns with energy conservation principles.
A charged lipid vesicle ($q=+4.0\times10^{-15}\ \text{C}$) is placed in a uniform electric field directed downward (−y). Gravity is negligible compared with electric effects. Based on the setup, which outcome is most likely for the direction of the electric force on the vesicle?
Zero, because a vesicle is electrically neutral overall by definition.
Downward (−y), because the force on a positive charge is along $\vec E$.
Perpendicular to −y, because uniform fields cause sideways drift only.
Upward (+y), because positive charges move toward higher potential.
Explanation
This question assesses understanding of electrostatics and electric fields in the context of force direction on a charged vesicle. Electrostatics involves the study of forces between charges, described by Coulomb's law and the concept of electric fields. In this setup, the electric field is uniform downward, affecting the positively charged vesicle. Choice B is correct because for q > 0, the force is along E, downward. Choice A is incorrect because it suggests upward motion, confusing potential with force direction. When analyzing electric fields, always use F = qE for direction; ensure reasoning aligns with field pointing from high to low potential.
A lab setup uses two large plates to create a uniform electric field. The potential difference is doubled while the plate separation is held constant. What would be the expected effect on the electric field magnitude between the plates (assume edge effects are negligible)?
It is unchanged, because $E$ depends only on plate area.
It becomes nonuniform, because higher voltage creates magnetic fields.
It halves, because capacitance increases with voltage.
It doubles, because $E\approx \Delta V/d$.
Explanation
This question assesses understanding of electrostatics and electric fields in the context of field magnitude changes in a parallel-plate setup. Electrostatics involves the study of forces between charges, described by Coulomb's law and the concept of electric fields. In this setup, the electric field is uniform, approximated by E ≈ ΔV/d. Choice A is correct because doubling ΔV at constant d doubles E. Choice B is incorrect because it confuses field with capacitance effects. When analyzing electric fields, always apply E = ΔV/d for uniform cases; ensure reasoning aligns with proportional changes.
In a simplified model of a membrane patch, a uniform electric field points from the extracellular side toward the cytosol. A chloride ion ($q=-1.6\times10^{-19}\ \text{C}$) is in the field. Which statement best describes the interaction of the charge with the field direction?
The ion experiences a magnetic force because the field induces circular motion.
The ion experiences a force in the same direction as the field because it is an anion.
The ion experiences a force opposite the field because $q<0$ in $\vec F=q\vec E$.
The ion experiences no force because electric fields do not act on ions in solution.
Explanation
This question assesses understanding of electrostatics and electric fields in the context of force direction on an anion in a membrane model. Electrostatics involves the study of forces between charges, described by Coulomb's law and the concept of electric fields. In this setup, the electric field is uniform across the membrane, affecting the chloride ion. Choice B is correct because for q < 0, the force is opposite E. Choice A is incorrect because it assumes the same direction, valid for cations but not anions. When analyzing electric fields, always factor in charge sign; ensure reasoning aligns with F = qE vectorially.
A uniform electric field points to the right (+x). Two test particles are released from rest: Particle 1 has $q=+e$ and Particle 2 has $q=-e$ (with $e=1.6\times10^{-19}\ \text{C}$). Which statement is most consistent with their initial accelerations (ignore gravity and drag)?
Both accelerate to the left because the field attracts all charges to lower potential.
Neither accelerates because uniform electric fields cannot change velocity.
Particle 1 accelerates right and Particle 2 accelerates left, because $\vec F=q\vec E$.
Both accelerate to the right because acceleration depends on field direction only.
Explanation
This question assesses understanding of electrostatics and electric fields in the context of acceleration directions for oppositely charged particles. Electrostatics involves the study of forces between charges, described by Coulomb's law and the concept of electric fields. In this setup, the electric field is uniform along +x, affecting particles of opposite charges. Choice C is correct because positive accelerates along E, negative opposite. Choice A is incorrect because it ignores charge sign effects on direction. When analyzing electric fields, always apply F = qE for each charge; ensure reasoning aligns with vector directions.
In a cell-sorting device, a uniform electric field is increased to enhance deflection of charged cells. For a cell modeled as a point particle with fixed net charge $q$ and mass $m$, moving in a region where only a uniform electric field acts, what would be the expected effect of increasing the electric field strength $E$ on the magnitude of the cell’s acceleration?
It decreases, because stronger fields reduce electric force by screening
It increases linearly, because $a=|q|E/m$
It becomes perpendicular to motion, because electric fields act like magnetic fields at high strength
It is unchanged, because acceleration depends only on mass
Explanation
This question assesses understanding of electrostatics and electric fields in the context of charged cell sorting. Electrostatics involves Newton's second law applied to charged particles, where the acceleration is given by a = F/m = qE/m for a charge in an electric field. In this setup, a cell with fixed charge q and mass m experiences only the electric force. Choice A is correct because the acceleration magnitude a = |q|E/m increases linearly with electric field strength E, since q and m are constant. Choice B is incorrect because it suggests that stronger fields reduce force through screening, which contradicts the direct proportionality F = qE and confuses screening effects with field strength effects. When analyzing charged particle dynamics, always apply Newton's second law with the electric force to find that acceleration is directly proportional to field strength.
A capacitor-based sensor is constructed to detect small changes in plate spacing caused by tissue swelling. The sensor is operated at fixed charge $Q$ on the plates (isolated capacitor). If the plate separation $d$ increases slightly while $Q$ is held constant, what would be the expected effect on the electric field magnitude between the plates (edge effects neglected)?
It reverses direction, because increasing $d$ changes the sign of $Q$
It decreases, because $E=\Delta V/d$ and $\Delta V$ must remain constant
It remains approximately constant, because $E\approx Q/(\varepsilon_0 A)$ for parallel plates at fixed $Q$
It increases, because $E=Q/(\varepsilon_0 A)$ and is independent of $d$
Explanation
This question assesses understanding of electrostatics and electric fields in the context of a capacitor-based sensor at fixed charge. Electrostatics involves the relationship between charge, field, and geometry in capacitors, where for parallel plates E = σ/ε₀ = Q/(ε₀A) when charge is fixed. In this setup, the capacitor holds constant charge Q while the plate separation d changes. Choice C is correct because the electric field between parallel plates depends on the surface charge density σ = Q/A, giving E = Q/(ε₀A), which is independent of the plate separation d. Choice B is incorrect because it applies the formula E = ΔV/d, which is valid but ΔV changes when d changes at fixed Q, keeping E constant. When analyzing capacitors at fixed charge, remember that E depends on charge density, not separation, making it invariant to spacing changes.
In a benchtop setup, a charged lipid vesicle with charge $q=+4.0\times10^{-19}\ \mathrm{C}$ is released from rest between large parallel plates producing a uniform electric field of $E=5.0\times10^{3}\ \mathrm{V/m}$ directed downward. Which outcome is most likely for the vesicle’s initial motion (neglect buoyancy and drag)?
It remains at rest, because $\mathrm{V/m}$ cannot be used to compute force
It moves horizontally, because $\vec{E}$ produces a perpendicular magnetic force
It accelerates downward, because the electric force on a positive charge is in the direction of $\vec{E}$
It accelerates upward, because positive charges move opposite $\vec{E}$
Explanation
This question assesses understanding of electrostatics and electric fields in the context of a charged vesicle between parallel plates. Electrostatics involves the study of forces on charges in electric fields, with F = qE determining both magnitude and direction of the force. In this setup, the vesicle has positive charge and the electric field points downward. Choice B is correct because a positive charge experiences a force in the same direction as the electric field, so with E pointing downward, the vesicle accelerates downward. Choice A is incorrect because it states that positive charges move opposite to the field direction, which is actually the behavior of negative charges. When analyzing motion of charged particles, always remember that positive charges follow the field direction while negative charges move opposite to it.
A researcher reports an electric field strength of $2.0\times10^{3}\ \text{V/m}$ in a tissue scaffold. Which unit is most consistent with electric field strength?
$\text{T}$
$\text{C/N}$
$\text{N/C}$
$\text{J/C}$
Explanation
This question assesses understanding of electrostatics and electric fields in the context of proper unit usage. Electrostatics involves the study of forces between charges, and electric field strength has specific equivalent units. In this setup, the researcher reports a field of 2.0 × 10³ V/m, and we need to identify the equivalent unit. Choice A is correct because N/C (newtons per coulomb) is the fundamental unit for electric field strength and is equivalent to V/m (volts per meter). Choice C is incorrect because J/C (joules per coulomb) is the unit for electric potential (voltage), not electric field. When working with electric fields, remember that E can be expressed as either N/C (force per unit charge) or V/m (potential difference per unit distance).
Two point charges are fixed: $q_1=-2.0\ \text{nC}$ and $q_2=-6.0\ \text{nC}$, separated by $r=0.30\ \text{m}$. Which statement best describes the interaction of the charges?
They repel because charges with the same sign repel.
They experience no force because both charges are negative.
They attract because their charges have the same sign but different magnitudes.
They repel due to magnetic forces created by stationary charges.
Explanation
This question assesses understanding of electrostatics and electric fields in the context of interactions between negative charges. Electrostatics involves the study of forces between charges, where like charges repel regardless of their magnitudes. In this setup, both charges are negative (-2.0 nC and -6.0 nC), making them like charges. Choice B is correct because charges with the same sign always repel according to Coulomb's law, regardless of their different magnitudes. Choice A is incorrect because it suggests same-sign charges can attract if magnitudes differ, which violates the fundamental principle that like charges repel. When analyzing charge interactions, the key factor is the sign: same signs repel, opposite signs attract, independent of magnitude differences.