Electrostatics

Electrostatics

Part 1 of 4 — Charge & Field Concepts

Electric Charge

• Charge comes in two types: positive and negative. Like charges repel, opposites attract.
• Charge is quantized: every charge is a whole-number multiple of the elementary charge $e = 1.6 \times 10^{-19}\;\text{C}$. An electron carries $-e$; a proton carries $+e$.
• Charge is conserved: rubbing a balloon on hair transfers electrons, it never creates or destroys charge.

Conductors vs. Insulators

• Conductors (metals, salt water, extracellular fluid): charges move freely. In electrostatic equilibrium the field INSIDE a conductor is zero and excess charge sits on the surface.
• Insulators (glass, rubber, the lipid bilayer of a cell membrane): charges stay put. A membrane can therefore hold separated charge on its two faces.

Coulomb's Law (Qualitative Picture)

$F = k\frac{q_1 q_2}{r^2}$ with $k = 8.99 \times 10^9\;\text{N}\cdot\text{m}^2/\text{C}^2$

• Inverse-square law, exactly like gravity — but electric force can attract OR repel, and it is enormously stronger than gravity for particles.
• Both charges enter symmetrically: by Newton's third law, each charge feels the same magnitude of force.

Electric Field

$\vec{E} = \frac{\vec{F}}{q}$ — the force per unit positive test charge, in N/C (equivalently V/m).

• Field lines point AWAY from positive charges and TOWARD negative charges.
• Density of field lines indicates field strength.
• A positive charge placed in a field feels a force along $\vec{E}$; a negative charge feels a force opposite $\vec{E}$.

Potential vs. Potential Energy — Keep Them Straight

• Positive charges spontaneously "fall" from high $V$ to low $V$.
• Negative charges spontaneously move from low $V$ to high $V$ — both directions LOWER the potential energy $U = qV$.

Equipotentials and Dipoles

• Equipotential surfaces are everywhere perpendicular to field lines; moving a charge along one takes zero work from the electric force.
• An electric dipole (equal $+q$ and $-q$ separated by a small distance — think of a polar molecule like water) feels zero NET force in a uniform field but a torque that rotates it into alignment with the field.

Worked Example — Ratio Reasoning with Coulomb's Law

Two ions attract each other with force $F$. The distance between them is then doubled and one ion's charge is replaced by an ion of three times the charge. What is the new force?

Step 1 — Write the proportionality. $F \propto \dfrac{q_1 q_2}{r^2}$.

Step 2 — Apply each change as a factor.

• Tripling one charge multiplies $F$ by $3$.
• Doubling $r$ multiplies $F$ by $\dfrac{1}{2^2} = \dfrac{1}{4}$.

Step 3 — Combine.

$F_{new} = F \times 3 \times \frac{1}{4} = \frac{3}{4}F$

The new force is $0.75F$, still attractive. This factor-by-factor style is how the MCAT expects you to handle Coulomb's law — almost never with a full plug-and-chug.

Worked Example — Which Way Does the Charge Move?

A chloride ion ($\text{Cl}^-$, charge $-e$) sits in a region where the potential increases from left ($V = -20\;\text{mV}$) to right ($V = +40\;\text{mV}$). Which way does the electric force push it?

Step 1 — Use energy. $U = qV$. With $q < 0$, the ion's energy is LOWEST where $V$ is HIGHEST.

Step 2 — Conclusion. The force pushes the chloride ion to the right, toward $+40\;\text{mV}$. Equivalently: $\vec{E}$ points from high to low potential (right to left here), and a negative charge feels force opposite $\vec{E}$ — again, to the right.

Charge & Field Concepts 🎯

Key Takeaways — Part 1

• Charge is quantized (multiples of $e = 1.6 \times 10^{-19}\;\text{C}$) and conserved; friction moves electrons only
• Conductors: zero interior field, charge on surface; insulators (like membranes) hold charge in place
• Coulomb: $F \propto q_1 q_2/r^2$ — solve MCAT problems with factor-by-factor ratio reasoning
• Field $\vec{E}$ is a vector (away from $+$, toward $-$); potential $V$ is a scalar; energy $U = qV$
• Positive charges fall toward low $V$; negative charges toward high $V$ — both lower $U$
• Uniform field + dipole = torque without net force

Electrostatics

Part 2 of 4 — The Quantitative Toolkit

Every electrostatics calculation on the MCAT comes from a short list of formulas. Learn which quantity each one produces and how it scales with $r$.

The Core Four (Point Charge $Q$, Distance $r$)

Memory hook: the two "pair" quantities ($F$, $U$) contain both charges; the two "field" quantities ($E$, $V$) contain one. The two energy-flavored quantities ($V$, $U$) go as $1/r$; the two force-flavored ones ($F$, $E$) go as $1/r^2$.

Connecting Formulas

• Force from field: $F = qE$
• Energy from potential: $U = qV$, and work done BY the field: $W = q(V_i - V_f) = -q\,\Delta V$
• Uniform field (parallel plates separated by $d$): $E = \dfrac{V}{d}$

Units Check

$1\;\text{V} = 1\;\text{J/C}$, and $1\;\text{N/C} = 1\;\text{V/m}$. If an answer choice's units don't work out, eliminate it.

Worked Example — Field and Potential of a Point Charge

A charge $Q = +5.0 \times 10^{-9}\;\text{C}$ sits at the origin. Find $E$ and $V$ at $r = 3.0\;\text{m}$. (Use $k \approx 9.0 \times 10^9\;\text{N}\cdot\text{m}^2/\text{C}^2$.)

Field:

$E = k\frac{Q}{r^2} = \frac{(9.0 \times 10^9)(5.0 \times 10^{-9})}{(3.0)^2} = \frac{45}{9} = 5.0\;\text{N/C}$, pointing away from $Q$.

Potential:

$V = k\frac{Q}{r} = \frac{(9.0 \times 10^9)(5.0 \times 10^{-9})}{3.0} = \frac{45}{3} = 15\;\text{V}$

Notice $V = E \cdot r$ for a point charge — a quick consistency check.

Worked Example — Work to Move a Charge

How much work must YOU do to push a charge $q = +2.0 \times 10^{-6}\;\text{C}$ slowly from a point at $100\;\text{V}$ to a point at $200\;\text{V}$?

Step 1 — The charge's potential energy changes by

$\Delta U = q\,\Delta V = (2.0 \times 10^{-6})(200 - 100) = 2.0 \times 10^{-4}\;\text{J}$

Step 2 — Moving slowly (no kinetic energy change), your work equals $\Delta U$:

$W_{you} = +2.0 \times 10^{-4}\;\text{J}$

The sign makes sense: pushing a positive charge toward higher potential is "uphill," so you do positive work while the field does $-2.0 \times 10^{-4}\;\text{J}$.

Worked Example — Parallel Plates

Two parallel plates $2.0\;\text{mm}$ apart are connected to a $6.0\;\text{V}$ battery. The field between them:

$E = \frac{V}{d} = \frac{6.0}{2.0 \times 10^{-3}} = 3.0 \times 10^{3}\;\text{V/m}$

Uniform everywhere between the plates, pointing from the $+$ plate to the $-$ plate.

The Electrostatics Toolkit 🎯

Key Takeaways — Part 2

• Pair quantities ($F$, $U$) use both charges; field quantities ($E$, $V$) use one
• Force-flavored: $F$, $E \propto 1/r^2$. Energy-flavored: $U$, $V \propto 1/r$
• Bridges: $F = qE$, $U = qV$, and $E = V/d$ between parallel plates
• Work by the field: $W = -q\,\Delta V$; along equipotentials $W = 0$
• Always convert mm and $\mu\text{C}$ to SI before plugging in — unit slips are the top distractor trap

Electrostatics

Part 3 of 4 — Membranes & Electrophoresis

Electrostatics is the MCAT's favorite bridge between physics and biology. Two applications dominate: the cell membrane as a charged parallel-plate system, and gel electrophoresis.

The Membrane as a Parallel-Plate Capacitor

The lipid bilayer is an insulator about $5$–$10\;\text{nm}$ thick, with conducting salt solutions on both sides. At rest, the inside of a neuron sits about $-70\;\text{mV}$ relative to the outside.

• Excess negative charge lines the inner face; positive charge lines the outer face — just like capacitor plates.
• The field inside the membrane is essentially uniform: $E = V/d$.
• Because $d$ is tiny, this field is HUGE (as you'll compute below) even though the voltage is small.

Depolarization in Field Language

When sodium channels open, $\text{Na}^+$ flows inward — DOWN its electrical gradient (positive charge attracted to the negative interior) and down its concentration gradient. The membrane potential rises toward zero and beyond: depolarization. Restoring $-70\;\text{mV}$ (repolarization) requires the $\text{Na}^+/\text{K}^+$ pump to do work against the field.

Gel Electrophoresis

A uniform field $E$ is applied across a gel. A molecule with charge $q$ feels a constant force $F = qE$.

• DNA's phosphate backbone makes it uniformly NEGATIVE, so DNA migrates toward the POSITIVE electrode (the anode).
• Because drag in the gel grows with size, smaller fragments travel farther in the same time — the field supplies the same force per charge, the gel does the size-sorting.
• In SDS-PAGE, the SDS coating gives proteins a roughly uniform negative charge-to-mass ratio, so separation is again by size.

Worked Example — Field Inside a Membrane

A neuron's membrane is $7.0\;\text{nm}$ thick with a resting potential of $70\;\text{mV}$ across it. Estimate the electric field inside the membrane.

Step 1 — Convert to SI. $V = 70\;\text{mV} = 7.0 \times 10^{-2}\;\text{V}$ and $d = 7.0 \times 10^{-9}\;\text{m}$.

Step 2 — Uniform-field formula.

$E = \frac{V}{d} = \frac{7.0 \times 10^{-2}}{7.0 \times 10^{-9}} = 1.0 \times 10^{7}\;\text{V/m}$

Ten million volts per meter — one of the strongest steady fields anywhere in biology, sustained by a $9\;\text{V}$-battery-free cell. This is why voltage-gated channels can respond so sharply: their charged gating segments sit inside an enormous field.

Worked Example — Energy to Move One Ion

How much energy does it take to move one $\text{Ca}^{2+}$ ion (charge $+2e$) across a membrane against a potential difference of $70\;\text{mV}$?

Step 1 — $q = 2e = 2(1.6 \times 10^{-19}) = 3.2 \times 10^{-19}\;\text{C}$

Step 2 —

$\Delta U = q\,\Delta V = (3.2 \times 10^{-19})(7.0 \times 10^{-2}) = 2.24 \times 10^{-20} \approx 2.2 \times 10^{-20}\;\text{J}$

Compare: hydrolyzing one ATP releases roughly $5 \times 10^{-20}\;\text{J}$ under cellular conditions — the same order of magnitude, which is why one ATP can power the transport of a few ions.

Membranes & Electrophoresis 🎯

Key Takeaways — Part 3

• Membrane = parallel-plate capacitor: insulating bilayer, conducting fluids, $E = V/d$
• $70\;\text{mV}$ over $7\;\text{nm}$ gives $E = 1 \times 10^{7}\;\text{V/m}$ — tiny voltage, giant field
• Depolarization shrinks $|V|$ and hence $E$; pumps do work $q\,\Delta V$ per ion to restore it
• Energy per ion crossing: $\Delta U = q\,\Delta V \approx 10^{-20}\;\text{J}$, same scale as ATP
• Electrophoresis: uniform field, $F = qE$; DNA (negative) runs to the anode; the gel sorts by size

Electrostatics

Part 4 of 4 — MCAT Integration

MCAT electrostatics questions rarely test one formula in isolation. They combine energy conservation, scaling, and sign logic — often wrapped in a biological or lab passage. Here is the playbook.

Strategy 1 — Energy Conservation Beats Kinematics

A charge accelerated from rest through a potential difference $\Delta V$ gains kinetic energy

$KE = |q\,\Delta V|$

no matter the path, the field geometry, or the distance. If the question gives you a voltage and asks for speed or energy, skip the field entirely.

Strategy 2 — Superposition: Fields Add as Vectors, Potentials as Numbers

• Midway between two EQUAL positive charges: fields cancel ($E = 0$) but potentials ADD ($V = 2kQ/r$, definitely not zero).
• Midway between a $+Q$ and a $-Q$: potentials cancel ($V = 0$) but the fields point the SAME way (from $+$ toward $-$) and add.

$E = 0$ and $V = 0$ are independent facts — the exam loves this asymmetry.

Strategy 3 — Track What's Held Constant

For a charged parallel-plate capacitor:

• Battery still connected ($V$ fixed): pulling plates apart decreases $E = V/d$.
• Battery disconnected ($Q$ fixed): the field, set by the charge on the plates, stays the same, so $V = Ed$ INCREASES as plates separate — you did work on the system.

Strategy 4 — Same Field, Different Particles

In the same field, a proton and an electron feel forces of EQUAL magnitude ($F = eE$) in opposite directions, but the electron's acceleration is about $1836$ times larger because $a = F/m$ and its mass is that much smaller.

Worked Example — Electron Gun (Energy Method)

An electron starts from rest and is accelerated through a potential difference of $500\;\text{V}$. Find its final kinetic energy and estimate its speed. ($m_e = 9.1 \times 10^{-31}\;\text{kg}$)

Step 1 — Kinetic energy from the voltage.

$KE = e\,\Delta V = (1.6 \times 10^{-19})(500) = 8.0 \times 10^{-17}\;\text{J}$

Step 2 — Speed from $KE = \frac{1}{2}mv^2$.

$v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2(8.0 \times 10^{-17})}{9.1 \times 10^{-31}}} = \sqrt{1.76 \times 10^{14}} \approx 1.3 \times 10^{7}\;\text{m/s}$

About $4\%$ of light speed from only $500\;\text{V}$ — electrons are that light. Note the electron accelerates from LOW potential toward HIGH potential (negative charge).

Worked Example — Midpoint Superposition

Two charges of $+4.0\;\text{nC}$ each are placed $2.0\;\text{m}$ apart. Find $E$ and $V$ at the midpoint.

Field: each charge produces a field of magnitude $kQ/r^2$ at the midpoint ($r = 1.0\;\text{m}$), but the two fields point in opposite directions and cancel exactly:

$E_{mid} = 0$

Potential: scalars add,

$V_{mid} = 2 \times \frac{kQ}{r} = 2 \times \frac{(9.0 \times 10^9)(4.0 \times 10^{-9})}{1.0} = 2 \times 36 = 72\;\text{V}$

A test charge placed exactly at the midpoint feels no force, yet has potential energy $qV = 72q$ joules — a perfect illustration that $E$ and $V$ carry independent information.

MCAT Integration: Electrostatics 🎯

Key Takeaways — Part 4

• Voltage-to-energy shortcut: $KE = |q\,\Delta V|$, path-independent — the fastest route to speed questions
• Superposition: fields add as vectors (can cancel), potentials add as signed numbers (can cancel) — but never assume one is zero because the other is
• Capacitor logic: battery connected fixes $V$; disconnected fixes $Q$ (and hence $E$) — decide FIRST what's constant
• Equal charges in equal fields feel equal forces; lighter particles accelerate more
• Negative charges spontaneously run "uphill" in $V$ — always convert to energy language when signs get confusing