Comprehensive Study Notes on Grade 11 Electrostatics and Circuitry and Electrostatics

Electrostatic Forces and Coulomb's Law

• Interactions Between Point Charges: The magnitude of the electrostatic force between two point charges is determined by Coulomb's Law, stated as $F = k \frac{|q_1 q_2|}{r^2}$. The direction is attractive for opposite charges and repulsive for like charges.     * Force on q2 from Fixed Coordinates:         * Particle 1: $q_1 = +3.0\, \mu C$ at coordinates $(3.5\, cm, 0.50\, cm)$.         * Particle 2: $q_2 = -4.0\, \mu C$ at coordinates $(-2.0\, cm, 1.5\, cm)$.         * Calculation of net force requires finding the distance vector between $(x_1, y_1)$ and $(x_2, y_2)$, calculating the individual vector components of the force, and determining the resultant magnitude and direction.     * Equilibrium of Three Charges: To make the net force on $q_2$ zero by adding a third charge $q_3 = +4.0\, \mu C$, the charge $q_3$ must be placed such that the force vector $\mathbf{F}_{32}$ is equal in magnitude and opposite in direction to $\mathbf{F}_{12}$.     * Ball Equilibrium in an Electric Field:         * A ball with mass $m = 0.010\, kg$ and charge $q$ is suspended by a light string in a uniform electric field $E = 5000\, N/C$.         * The string makes an angle of $\theta = 37^{\circ}$ with the vertical.         * The Free Body Diagram (FBD) includes: Tension ($T$) acting along the string, Gravity ($mg$) acting downward, and Electrostatic Force ($F_e = qE$) acting horizontally.         * Equations for equilibrium:             * $T \cos(37^{\circ}) = mg$             * $T \sin(37^{\circ}) = qE$         * Dividing these yields $\tan(37^{\circ}) = \frac{qE}{mg}$. From this, the magnitude of charge $q$ can be calculated. The sign of the charge is determined by the direction of displacement relative to the field direction: if the ball moves in the direction of the field, it is positive; if opposite, it is negative.     * Resultant Field in Equilateral Triangles: For charges $+q$ and $-2q$ at vertices $B$ and $C$ of an equilateral triangle $ABC$, the electric field at vertex $A$ is the vector sum of $\mathbf{E}_B$ and $\mathbf{E}_C$.         * $|E_B| = k \frac{q}{a^2}$         * $|E_C| = k \frac{2q}{a^2}$         * The angle between the field vectors is $120^{\circ}$ because of the geometry of the equilateral triangle and the sign of the charges.     * Force Midpoint Cancellation: For two identical charges, the magnitude of the net electric field at the exact midpoint on the straight line connecting them is zero ($E_{net} = 0$), as the field vectors from each charge are equal in magnitude but opposite in direction.

Electric Fields, Flux, and Gauss's Law

• Electric Field Calculations:     * For point charges: $E = k \frac{Q}{r^2}$.     * Between parallel plates: $E = \frac{V}{d}$.
• Electric Flux ($\Phi$): Defined as the product of the electric field and the area through which it passes perpendicular to the field: $\Phi = E \cdot A \cdot \cos(\theta)$.     * Flux through a Sphere: A sphere of radius $r$ surrounding a charge creating a field of strength $E$ at distance $r$ has a total flux $\Phi = E \times 4\pi r^2$.     * Flux Through a Cube: According to Gauss's Law, if a charge $q$ is at the center of a cube, the total flux through the cube is $\frac{q}{\epsilon_0}$. The flux through a single face is $\frac{q}{6\epsilon_0}$.     * Vector Field Flux: If the field $\mathbf{E} = 5\mathbf{i} + 2\mathbf{j}$ hits a $10\, m^2$ area in the $y-z$ plane, only the component perpendicular to the plane ($E_x=5$) contributes to flux. $\Phi = E_x \times A = 5 \times 10 = 50\, V\cdot m$.     * Hemispherical Flux: For a hemisphere in a field $E$ pointing into the base, the flux through the curved side must equal the flux through the circular base to satisfy Gauss's Law (if no charge is enclosed), resulting in $\Phi = \pi R^2 E$.
• Induced Charges on Conductors:     * A solid conducting sphere ($R_1 = 5.0\, cm$, $Q_1 = +4.0\, nC$) inside a hollow uncharged conducting shell ($R_2 = 10.0\, cm, R_3 = 15.0\, cm$) results in induced charges.     * $\sigma_{inner}$ on the inner surface ($r=R_2$) is based on the charge $-Q_1$ required to cancel the internal field. $\sigma_{inner} = \frac{-Q_1}{4\pi R_2^2}$.     * $\sigma_{outer}$ on the outer surface ($r=R_3$) must be $+Q_1$ to maintain the shell's neutrality. $\sigma_{outer} = \frac{Q_1}{4\pi R_3^2}$.

Electric Potential and Energy

• Potential Formulas:     * Point charge: $V = k \frac{Q}{r}$.     * Relation to field: $\Delta V = -E \cdot d$ or $E = -\frac{dV}{dx}$.
• Work Done by Field: $W = q \Delta V$. Moving a charge along an equipotential surface requires zero work ($W = 0$).     * Example: Moving $5\, \mu C$ from $20\, V$ to $70\, V$: $W = q(V_f - V_i) = 5\, \mu C \times 50\, V = 250\, \mu J$.
• Potential Energy of a System: The total energy of a configuration of charges (e.g., three charges at triangle corners) is the sum of the potential energies of every unique pair: $U_{total} = \sum k \frac{q_i q_j}{r_{ij}}$.
• Alpha Particle Acceleration: An alpha particle ($q = +2e$) moving through a potential difference of $1,000,000\, V$ gains kinetic energy $K = qV = 2e \times 10^6\, V = 2\, MeV$.
• Potential Inside a Conductor: A solid metal ball charged to $100\, V$ has a constant potential throughout its volume. A voltmeter reading halfway between the center and the surface would show $100\, V$.
• Potential of a Dipole: The potential $V$ of a dipole varies with distance $r$ as $V \propto \frac{1}{r^2}$.
• Zero Field vs. Zero Potential: In a "Quiet Field" where the voltage is constant everywhere (e.g., $10\, V$), the electric field is zero because there is no potential gradient ($E = -\nabla V$).

Capacitance and Dielectrics

• Capacitance Fundamentals: $C = \frac{Q}{V}$. For a parallel plate capacitor: $C = \frac{\epsilon_0 A}{d}$.
• Dielectric Insertion:     * When a dielectric with constant $\kappa$ is inserted, capacitance increases: $C = \kappa C_0$.     * Constant Battery Connection: If the battery remains connected ($V$ is constant), the charge increases ($Q = CV$). The battery does work ($W = \Delta Q \cdot V$) to push more charge onto the plates.     * Disconnected Capacitor: If the capacitor is disconnected ($Q$ is constant) and plates are pulled apart ($d$ increases), capacitance decreases. The stored energy ($U = \frac{1}{2} \frac{Q^2}{C}$) increases because mechanical work is done to overcome the attraction between plates.
• Energy and Charge Calculations:     * Example: Air capacitor $C_0 = 12.0\, nF$, $V_0 = 50.0\, V$, dielectric $\kappa = 3.5$.     * $\Delta Q = Q_{final} - Q_{initial} = (\kappa C_0 V) - (C_0 V) = V C_0(\kappa - 1)$.     * $\Delta U = \frac{1}{2} C_{final} V^2 - \frac{1}{2} C_0 V^2$.
• Coaxial Cable Capacitance: For inner radius $a$ and outer radius $b$, capacitance per unit length is $\frac{C}{L} = \frac{2\pi \epsilon_0}{\ln(b/a)}$.

Current, Resistance, and DC Circuits

• Resistance and Resistivity: $R = \rho \frac{L}{A}$.     * Aluminum wire example: $L = 10\, m$, $A = 5\, mm^2 = 5 \times 10^{-6}\, m^2$, $\rho = 1.8 \times 10^{-8}\, \Omega\cdot m$.     * $R = 1.8 \times 10^{-8} \times \frac{10}{5 \times 10^{-6}} = 0.036\, \Omega$.     * Resistance is an intrinsic property based on geometry and material; it does not change when the current varies (unless temperature changes).
• Current Density ($J$) and Drift Velocity ($v_d$):     * $J = \frac{I}{A} = nqv_d$.     * Copper wire example: Square cross-section length $2.0\, mm$, current $10\, A$, $n = 8 \times 10^{28}\, m^{-3}$.     * $A = (2.0 \times 10^{-3})^2 = 4 \times 10^{-6}\, m^2$.     * $J = \frac{10}{4 \times 10^{-6}} = 2.5 \times 10^6\, A/m^2$.     * Time for electron to travel length $L$: $t = \frac{L}{v_d}$.
• Kirchhoff’s Laws:     * Junction Rule: $\sum I_{in} = \sum I_{out}$.     * Loop Rule: $\sum V = 0$ around any closed loop.
• Wheatstone Bridge:     * Balanced condition: $\frac{P}{Q} = \frac{R}{S}$.     * In balance, the potential at the two mid-bridge points is equal, and no current flows through the galvanometer ($I_g = 0$).
• Potentiometer: Used to compare resistances or EMFs. For resistances $R$ and $X$, the ratio is $\frac{X}{R} = \frac{l_2}{l_1}$, where $l$ is the balance point length.
• Temperature Coefficient: Insulators and semiconductors have a negative temperature coefficient of resistance (resistance decreases as temperature increases), unlike metals.

Motion of Charged Particles

• Acceleration in a Field: $a = \frac{F}{m} = \frac{qE}{m}$.
• Kinetic Energy Gains: When particles are accelerated through a potential difference $V$, they gain kinetic energy $\Delta K = qV$.     * An electron and a proton dropped in the same field and falling the same distance gain the same amount of kinetic energy (since $q$ is the same magnitude), but the electron will have a higher speed due to its smaller mass.
• Path of Particle: A charged particle projected with velocity perpendicular to a uniform electric field follows a parabolic arc, similar to projectile motion in a gravitational field.
• Deflection Distance: The vertical deflection $y$ of a particle with horizontal velocity $v_x$ traveling distance $L$ in field $E$ is $y = \frac{1}{2} a t^2 = \frac{1}{2} (\frac{qE}{m}) (\frac{L}{v_x})^2$.

Questions & Discussion

• Q: What happens if you failed to find a balance point on a potentiometer?     * A: This might happen if the EMF of the driver cell is less than the potential drop across the length of the wire or if the unknown potential difference is higher than the total potential drop of the potentiometer wire. One could increase the driver cell voltage or the wire length.
• Q: What is the relation between dielectric constants K, K1, and K2 for combined slabs?     * A: If slabs are in parallel (split by area), $K = \frac{K_1 + K_2}{2}$. If in series (split by thickness), the relation is $\frac{1}{K} = \frac{1}{2} (\frac{1}{K_1} + \frac{1}{K_2})$.
• Q: How does the charge move between two differently sized spheres connected by a wire?     * A: Negative charge flows from the smaller sphere to the larger sphere until the electric potential at the surface of each sphere is the same ($V_1 = V_2$), which is the condition for electrostatic equilibrium.
• Q: What happens to a soap bubble if it is given a negative charge?     * A: Its radius increases due to the mutual repulsion of the negative charges distributed on its surface.
• Q: What is the vibe at the center of a square with two positive and two negative charges diagonal to each other?     * A: There is no net voltage ($V = 0$) because the potentials from opposite corners cancel out, but the field forces are present and pushing toward the negative charges.