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Physics: Electricity & Optics

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Physics: Electricity & Optics - Complete Interactive Lesson

Part 1: Electrostatics & Coulombs Law

Physics: Electricity, Magnetism & Optics

Part 1 of 7 — Electrostatics

Coulomb's Law

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

Like charges repel, opposite charges attract
Force is proportional to $1/r^2$ (inverse square law)
The elementary charge is $e = 1.6 \times 10^{-19}\;\text{C}$

Electric Field

$\vec{E} = \frac{F}{q} = k\frac{Q}{r^2}$

Points AWAY from positive charges, TOWARD negative charges
Units: N/C or V/m
Force on a charge placed in a field: $F = qE$

Electric Potential (Voltage)

$V = k\frac{Q}{r}$ — a scalar, so contributions from multiple charges simply add

Electric Potential Energy

$U = k\frac{q_1 q_2}{r} = qV$

Positive charges move from high $V$ to low $V$ spontaneously
Negative charges move from low $V$ to high $V$ spontaneously

Equipotential Insight

Moving along an equipotential surface requires no work by the electric field because $\Delta V = 0$ .

Worked Example — Coulomb's Law in a Membrane

Two ions carrying charges $q_1 = +2e$ and $q_2 = -e$ sit apart across a cell membrane. What is the magnitude of the electrostatic force between them?

Electrostatics 🎯

Key Takeaways — Part 1

Coulomb's law: $F \propto q_1 q_2/r^2$ , same inverse-square form as gravity
E-field points away from , toward ; force on a charge is

Part 2: Electric Circuits

Physics: Electricity, Magnetism & Optics

Part 2 of 7 — Circuits (HIGH YIELD)

Ohm's Law

$V = IR$ — voltage equals current times resistance

Kirchhoff's Laws

Junction Rule: Current in = current out ( $\sum I_{in} = \sum I_{out}$ )

Part 3: Magnetism & EM Induction

Physics: Electricity, Magnetism & Optics

Part 3 of 7 — Magnetism

Magnetic Force on a Moving Charge

$F = qvB\sin\theta$

Direction: Right-hand rule (fingers point from $\vec{v}$ to , thumb gives for a positive charge)

Part 4: Optics & Light

Physics: Electricity, Magnetism & Optics

Part 4 of 7 — Optics: Reflection & Refraction

Law of Reflection

$\theta_{incident} = \theta_{reflected}$ (angles measured from the normal)

Part 5: Nuclear Physics & Radioactivity

Physics: Electricity, Magnetism & Optics

Part 5 of 7 — Lenses & Mirrors

Thin Lens / Mirror Equation

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Part 6: Electrochemistry

Physics: Electricity, Magnetism & Optics

Part 6 of 7 — Electromagnetic Spectrum & Light

The EM Spectrum (increasing frequency / decreasing wavelength)

Radio → Microwave → Infrared → Visible → Ultraviolet → X-ray → Gamma

$c = f\lambda$ where $c = 3 \times 10^8\;\text{m/s}$

Part 7: Review & MCAT Practice

Physics: Electricity, Magnetism & Optics

Part 7 of 7 — Atomic & Nuclear Physics

Atomic Models on the MCAT

Bohr model: electrons occupy quantized orbits with $E_n = -13.6/n^2\;\text{eV}$ (for hydrogen)
A photon is emitted when an electron drops levels:

r = 2.0 \times 10^{-9}\;\text{m}

Step 1 — Convert charges to coulombs.

$q_1 = 2(1.6 \times 10^{-19}) = 3.2 \times 10^{-19}\;\text{C}$ and $q_2 = 1.6 \times 10^{-19}\;\text{C}$ (use magnitudes for force size).

Step 2 — Apply Coulomb's law.

$F = k\frac{q_1 q_2}{r^2} = (8.99 \times 10^9)\frac{(3.2 \times 10^{-19})(1.6 \times 10^{-19})}{(2.0 \times 10^{-9})^2}$

Step 3 — Evaluate the pieces. The numerator product of charges is $5.12 \times 10^{-38}$ and $r^2 = 4.0 \times 10^{-18}\;\text{m}^2$ , so

$F = (8.99 \times 10^9)\frac{5.12 \times 10^{-38}}{4.0 \times 10^{-18}} \approx 1.2 \times 10^{-10}\;\text{N}$

The force is attractive (opposite signs). On the MCAT, notice you rarely need an exact decimal — estimating powers of ten and confirming the direction (attractive vs. repulsive) is usually enough.

V = kQ/r

is a scalar (not a vector) — easier to add than fields

Potential energy:

U = kq_1q_2/r = qV

, and

U \propto 1/r

(force

\propto 1/r^2

)

Loop Rule: Total voltage change around any closed loop = 0 (

\sum V = 0

)

Series vs. Parallel Resistors

Configuration	Resistance	Current	Voltage
Series	$R_T = R_1 + R_2 + ...$	Same through each	Divides
Parallel	$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ...$

$P = IV = I^2R = \frac{V^2}{R}$ — know all three forms

$C = \frac{Q}{V}$ and stored energy $U = \frac{1}{2}CV^2$

Series: $\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2}$ (OPPOSITE of resistors!)
Parallel: $C_T = C_1 + C_2$

Capacitors resist instantaneous voltage change, which is why they smooth signals and set charging/discharging time constants $\tau = RC$ in physiology instrumentation contexts.

Worked Example — Series/Parallel Combination

A $12\;\text{V}$ battery connects to a $4\;\Omega$ resistor in series with a parallel pair of $6\;\Omega$ and $3\;\Omega$ resistors. Find the total current the battery supplies and the power it delivers.

Step 1 — Reduce the parallel pair.

$\frac{1}{R_P} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$ , so $R_P = 2\;\Omega$ .

Step 2 — Add the series resistor.

$R_T = 4 + R_P = 4 + 2 = 6\;\Omega$ .

Step 3 — Apply Ohm's law for total current.

$I = \frac{V}{R_T} = \frac{12}{6} = 2\;\text{A}$ .

Step 4 — Power delivered by the battery.

$P = IV = (2)(12) = 24\;\text{W}$ (equivalently $I^2 R_T = 4 \times 6 = 24\;\text{W}$ ).

The MCAT loves this pattern: collapse parallel groups first, then treat the rest as a simple series chain.

Key Takeaways — Part 2

Series: same current, voltages add, $R_{total}$ increases
Parallel: same voltage, currents add, $R_{total}$ decreases
Capacitors add OPPOSITE to resistors (parallel: $C$ adds; series: $1/C$ adds)
$P = IV = I^2R = V^2/R$ — pick the form that matches your known quantities
Strategy: collapse parallel groups first, then treat the rest as a series chain

Force is PERPENDICULAR to both velocity and field

Stationary charges feel NO magnetic force (

v = 0

)

Magnetic force does NO work, because it is always perpendicular to velocity

Circular Motion in a Magnetic Field

A charge moving perpendicular to $\vec{B}$ follows a circle. Setting magnetic force equal to centripetal force:

$qvB = \frac{mv^2}{r}$ which rearranges to $r = \frac{mv}{qB}$

Force on a Current-Carrying Wire

$F = ILB\sin\theta$ where $L$ is the length of wire in the field

Electromagnetic Induction (Faraday's Law)

$\varepsilon = -\frac{\Delta\Phi_B}{\Delta t}$ where magnetic flux $\Phi_B = BA\cos\theta$

A changing magnetic flux induces an EMF (voltage)
Lenz's Law: the induced current opposes the change that caused it

Flux can change because the field strength, the loop area, or the orientation relative to the field changes.

Worked Example — Radius in a Mass Spectrometer

A proton ( $m = 1.67 \times 10^{-27}\;\text{kg}$ , $q = 1.6 \times 10^{-19}\;\text{C}$ ) enters a $0.50\;\text{T}$ field perpendicular to its velocity at $v = 2.0 \times 10^{6}\;\text{m/s}$ . What is the radius of its circular path?

Step 1 — Use the circular-motion result.

$r = \frac{mv}{qB}$

Step 2 — Substitute the values.

$r = \frac{(1.67 \times 10^{-27})(2.0 \times 10^{6})}{(1.6 \times 10^{-19})(0.50)}$

Step 3 — Evaluate. Numerator $= 3.34 \times 10^{-21}$ ; denominator $= 8.0 \times 10^{-20}$ .

$r = \frac{3.34 \times 10^{-21}}{8.0 \times 10^{-20}} \approx 0.042\;\text{m} \approx 4.2\;\text{cm}$

This is exactly how a mass spectrometer separates ions: heavier or faster ions curve with a larger radius, while a stronger field $B$ tightens the curve.

Key Takeaways — Part 3

Magnetic force: $F = qvB\sin\theta$ (zero when $\vec{v}$ is parallel to $\vec{B}$ )
Magnetic force does NO work (always perpendicular to velocity)
Right-hand rule for direction: point fingers from $\vec{v}$ to $\vec{B}$ , thumb gives
Circular path radius: $r = mv/(qB)$ — the basis of the mass spectrometer
Faraday: changing flux induces EMF; Lenz: the induced current opposes the change

Snell's Law (Refraction)

$n_1\sin\theta_1 = n_2\sin\theta_2$

$n$ = index of refraction, defined by $n = c/v$ (always $\ge 1$ )
Light bends TOWARD the normal when entering a denser medium ( $n_2 > n_1$ )
Light bends AWAY from the normal when entering a less dense medium

Total Internal Reflection

$\sin\theta_c = \frac{n_2}{n_1}$ (requires $n_1 > n_2$ )

Only occurs going from a denser to a less dense medium
The angle of incidence must EXCEED the critical angle $\theta_c$
Applications: fiber optics, the sparkle of diamond, endoscopes

A higher refractive index means a lower light speed in that medium, since $n = c/v$ .

Worked Example — Snell's Law at a Water Surface

Light travels from air ( $n_1 = 1.00$ ) into water ( $n_2 = 1.33$ ), striking the surface at $\theta_1 = 30°$ from the normal. Find the refraction angle $\theta_2$ .

Step 1 — Write Snell's law.

$n_1\sin\theta_1 = n_2\sin\theta_2$

Step 2 — Solve for $\sin\theta_2$ .

$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{(1.00)\sin 30°}{1.33} = \frac{(1.00)(0.50)}{1.33} \approx 0.376$

Step 3 — Take the inverse sine.

$\theta_2 = \sin^{-1}(0.376) \approx 22°$

Since $\theta_2 < \theta_1$ , the ray bent TOWARD the normal — exactly what we expect when light enters a denser medium. On the MCAT, you usually only need this qualitative bending direction, not the exact angle.

Optics: Refraction 🎯

Key Takeaways — Part 4

Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$
Entering a denser medium → bend toward the normal (slower speed, $v = c/n$ )
Total internal reflection: only denser → less dense, beyond the critical angle
Frequency is conserved across a boundary; speed and wavelength change
All angles are measured from the NORMAL, not the surface

$m = -\frac{d_i}{d_o} = \frac{h_i}{h_o}$

$|m| > 1$ : enlarged; $|m| < 1$ : reduced
$m > 0$ : upright; $m < 0$ : inverted

Quantity	Positive	Negative
$d_o$	Object on the same side as incoming light	(Virtual object)
$d_i$	Image on the opposite side (real)	Same side as object (virtual)
$f$	Converging (convex lens / concave mirror)	Diverging (concave lens / convex mirror)

Concave mirror / Convex lens: Converging, $f > 0$
Convex mirror / Concave lens: Diverging, $f < 0$ , always produces a virtual, upright, reduced image

Power of a lens (diopters) is $P = 1/f$ with $f$ in meters. Always interpret the sign of $d_i$ and $m$ before choosing the image description.

Worked Example — Image from a Converging Lens

An object sits $d_o = 15\;\text{cm}$ in front of a converging lens with focal length $f = 10\;\text{cm}$ . Find the image distance and magnification, then describe the image.

Step 1 — Apply the thin-lens equation.

$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{10} - \frac{1}{15}$

Step 2 — Find a common denominator.

$\frac{1}{10} - \frac{1}{15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}$ , so .

Step 3 — Compute the magnification.

$m = -\frac{d_i}{d_o} = -\frac{30}{15} = -2$

Step 4 — Interpret. Positive $d_i$ means a REAL image (opposite side of the lens). Negative $m$ means INVERTED, and $|m| = 2$ means it is enlarged $2\times$ . So the image is real, inverted, and twice as tall — the kind of image a projector forms.

Lenses & Mirrors 🎯

Key Takeaways — Part 5

$1/f = 1/d_o + 1/d_i$ — works for both lenses and mirrors
Diverging elements ( $f < 0$ ): always virtual, upright, reduced
Sign of $d_i$ tells you real ( $+$ ) vs. virtual ( $-$ )
Sign of $m$ tells you inverted ( $-$ ) vs. upright ( $+$ ); $|m|$ gives the size ratio
Lens power in diopters: $P = 1/f$ with $f$ in meters

Red (700 nm) → Orange → Yellow → Green → Blue → Violet (400 nm)

$E = hf = \frac{hc}{\lambda}$ where $h = 6.63 \times 10^{-34}\;\text{J}\cdot\text{s}$

Higher frequency = higher energy = shorter wavelength

Photoelectric Effect

$KE_{max} = hf - \phi$

$\phi$ = work function (minimum energy to eject an electron)
Below the threshold frequency: NO electrons are ejected, regardless of intensity
Above the threshold: intensity sets the NUMBER of electrons, frequency sets their energy

Diffraction & Interference

Constructive (bright fringes): $d\sin\theta = n\lambda$
Destructive (dark fringes): $d\sin\theta = (n + \tfrac{1}{2})\lambda$

Below the threshold frequency, no electrons are emitted no matter how bright the light, because each single photon lacks the energy to overcome $\phi$ .

Worked Example — Photon Energy of Green Light

Green light has a wavelength of $\lambda = 500\;\text{nm} = 5.0 \times 10^{-7}\;\text{m}$ . What is the energy of a single green photon?

Step 1 — Use the photon-energy relation.

$E = \frac{hc}{\lambda}$

Step 2 — Substitute constants.

$E = \frac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{5.0 \times 10^{-7}}$

Step 3 — Evaluate. Numerator $= 1.99 \times 10^{-25}$ , so

$E = \frac{1.99 \times 10^{-25}}{5.0 \times 10^{-7}} \approx 4.0 \times 10^{-19}\;\text{J}$

Step 4 — Convert to electron-volts (optional MCAT shortcut). Dividing by $1.6 \times 10^{-19}\;\text{J/eV}$ gives about $2.5\;\text{eV}$ , a typical visible-photon energy. Remember the inverse relationship: shorter wavelength means higher energy per photon.

Light & Quantum 🎯

Key Takeaways — Part 6

$E = hf = hc/\lambda$ : higher frequency = higher energy = shorter wavelength
Photoelectric effect: threshold FREQUENCY matters for emission, not intensity
Intensity sets the NUMBER of photoelectrons; frequency sets their kinetic energy
EM spectrum order: Radio < Micro < IR < Visible < UV < X-ray < Gamma
Double-slit interference demonstrates the wave nature of light

E_{p h o t o n} = h f = E_{hi g h} - E_{l o w}

$^A_Z X$ where $A$ = mass number (protons + neutrons) and $Z$ = atomic number (protons)

Radioactive Decay Types

Type	Particle	Change in $A$	Change in $Z$
Alpha ( $\alpha$ )	$^4_2\text{He}$	$-4$	$-2$
Beta-minus ( $\beta^-$ )	Electron	$0$	$+1$
Beta-plus ( $\beta^+$ )	Positron	$0$	$-1$
Gamma ( $\gamma$ )	Photon	$0$	$0$

$N = N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}$

After $n$ half-lives: $N = N_0/2^n$ . Activity is proportional to the number of undecayed nuclei, so activity falls off with the same half-life behavior.

Worked Example — Half-Life Decay

A radioactive isotope used in a tracer study has a half-life of $t_{1/2} = 6\;\text{hours}$ . A sample starts with $8.0\;\text{mg}$ . How much remains after $18\;\text{hours}$ ?

Step 1 — Count the half-lives.

$n = \frac{t}{t_{1/2}} = \frac{18}{6} = 3$ half-lives.

Step 2 — Halve the amount once per half-life.

After 1: $8.0 \to 4.0\;\text{mg}$ . After 2: $4.0 \to 2.0\;\text{mg}$ . After 3: $2.0 \to 1.0\;\text{mg}$ .

Step 3 — Confirm with the formula.

$N = N_0 (1/2)^n = 8.0 \times (1/2)^3 = 8.0 \times \frac{1}{8} = 1.0\;\text{mg}$

So $1.0\;\text{mg}$ remains. The MCAT almost always uses a whole number of half-lives, so repeated halving is the fastest route.

Nuclear Physics 🎯

Physics E&M / Optics — Complete! ✅

Master circuits, optics (the lens/mirror equation), and nuclear decay — the most-tested physics topics. Remember the patterns: collapse parallel resistor groups, read the sign of $d_i$ for real vs. virtual images, and count whole half-lives for decay. The MCAT rewards understanding WHY over heavy computation.

(1.67×10−27)(2.0×106)

d_i = 30\;\text{cm}

Physics: Electricity & Optics

Series vs. Parallel Resistors

Power

Capacitors

RC Intuition

Worked Example — Series/Parallel Combination

Key Takeaways — Part 2

Circular Motion in a Magnetic Field

Force on a Current-Carrying Wire

Electromagnetic Induction (Faraday's Law)

Worked Example — Radius in a Mass Spectrometer

Key Takeaways — Part 3

Snell's Law (Refraction)

Total Internal Reflection

Worked Example — Snell's Law at a Water Surface

Key Takeaways — Part 4

Magnification

Sign Conventions

MCAT Must-Know

Worked Example — Image from a Converging Lens

Key Takeaways — Part 5

Visible Light

Photoelectric Effect

Diffraction & Interference

Worked Example — Photon Energy of Green Light

Key Takeaways — Part 6

Nuclear Notation

Radioactive Decay Types

Half-Life

Worked Example — Half-Life Decay

Physics E&M / Optics — Complete! ✅