FAD1022 CQ1: Rapid-Fire Drill Pack — Electrostatics

Objective: Master all Electrostatics concepts for CQ1.
Target: 3-4 min per problem (CQ1 style: 4 questions in 15 min).
Total problems: 32
Estimated time: ~120 min

Cheat Sheet (Memorize First)

Constants for CQ1

Constant Symbol Value
Coulomb constant $k$ $9.0 \times 10^{9}\ \text{N·m}^{2}\text{·C}^{-2}$
Permittivity of free space $\varepsilon_{0}$ $8.85 \times 10^{-12}\ \text{C}^{2}\text{·N}^{-1}\text{·m}^{-2}$
Elementary charge $e$ $1.60 \times 10^{-19}\ \text{C}$
Electron mass $m_{e}$ $9.11 \times 10^{-31}\ \text{kg}$

Core Formulas

Formula Description
$F = k \frac{q_1 q_2}{r^{2}}$ Coulomb's Law (magnitude)
$\vec{F} = q\vec{E}$ Force on charge in E-field
$E = k \frac{Q}{r^{2}}$ Electric field of point charge
$\vec{E}{\text{net}} = \sum \vec{E}{i}$ Superposition principle
$a = \frac{qE}{m}$ Acceleration in uniform E-field
$y = \frac{1}{2} a t^{2} = \frac{1}{2} \cdot \frac{qE}{m} \cdot \left(\frac{L}{v_{0}}\right)^{2}$ Deflection in parallel plates
$\theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right)$ Deflection angle

Quick Identification Rules

  • Force direction: Like charges repel, opposite charges attract
  • E-field direction: Points away from positive, toward negative
  • Electron in E-field: Accelerates opposite to E-field direction (negative charge)

Part A: Coulomb's Law — 1D Force Calculations (8 problems)

Target: 3 min per problem.

Set A1 — Two-Point Charges on a Line (4 problems)

  1. Two point charges $q_1 = +3.0\ \mu\text{C}$ and $q_2 = -5.0\ \mu\text{C}$ are separated by $r = 20\ \text{cm}$. Calculate the magnitude of the electrostatic force between them.

  2. Calculate the distance $r$ between two electrons if the electrostatic repulsion force equals $F = 2.3 \times 10^{-8}\ \text{N}$.

  3. Two identical charges experience a repulsive force of $F = 0.36\ \text{N}$ when separated by $r = 15\ \text{cm}$. Find the magnitude of each charge.

  4. Two charges $q_1 = +8.0\ \mu\text{C}$ and $q_2 = +2.0\ \mu\text{C}$ are placed $30\ \text{cm}$ apart. Where along the line between them is the electric field zero?

Score: ___/4

Set A2 — Three Charges in a Line (4 problems)

  1. Three charges lie on the x-axis: $q_1 = +2.0\ \mu\text{C}$ at $x = 0$, $q_2 = -4.0\ \mu\text{C}$ at $x = 30\ \text{cm}$, and $q_3 = +6.0\ \mu\text{C}$ at $x = 60\ \text{cm}$. Find the net force on $q_2$.

  2. For the same configuration as problem 5, find the net force on $q_1$.

  3. Three charges are arranged on a line: $q_A = +5.0\ \text{nC}$ at position $0$, $q_B = -3.0\ \text{nC}$ at position $10\ \text{cm}$, and $q_C = +4.0\ \text{nC}$ at position $25\ \text{cm}$. Calculate the net force on $q_B$.

  4. Two charges $q_1 = +4.0\ \mu\text{C}$ and $q_2 = -9.0\ \mu\text{C}$ are fixed $50\ \text{cm}$ apart. Where should a third charge $q_3 = +2.0\ \mu\text{C}$ be placed so that the net force on it is zero?

Score: ___/4

Part B: Coulomb's Law — 2D Force Calculations (6 problems)

Target: 4 min per problem.

Set B1 — Right-Angle Triangle Configurations (3 problems)

  1. Three charges form a right triangle: $q_1 = +3.0\ \mu\text{C}$ at the origin, $q_2 = -4.0\ \mu\text{C}$ at $(0, 30\ \text{cm})$, and $q_3 = +5.0\ \mu\text{C}$ at $(40\ \text{cm}, 0)$. Find the net force on $q_1$.

  2. Two charges $q_1 = +2.5\ \mu\text{C}$ and $q_2 = -6.0\ \mu\text{C}$ are placed at $(0, 0)$ and $(0.8\ \text{m}, 0)$ respectively. A third charge $q_3 = +4.0\ \mu\text{C}$ is at $(0.8\ \text{m}, 0.6\ \text{m})$. Find the net force on $q_3$.

  3. Three positive charges $q_1 = q_2 = q_3 = +2.0\ \mu\text{C}$ are placed at the vertices of a right triangle with legs $a = 30\ \text{cm}$ and $b = 40\ \text{cm}$. Find the net force on the charge at the right-angle vertex.

Score: ___/3

Set B2 — Equilateral and Isosceles Triangle (3 problems)

  1. Three charges form an equilateral triangle of side $a = 20\ \text{cm}$: $q_1 = +4.0\ \mu\text{C}$, $q_2 = +4.0\ \mu\text{C}$, and $q_3 = -4.0\ \mu\text{C}$. Find the net force on $q_3$.

  2. Three charges are at the corners of an equilateral triangle with side $L = 50\ \text{cm}$: $q_A = +6.0\ \mu\text{C}$, $q_B = +6.0\ \mu\text{C}$, and $q_C = -3.0\ \mu\text{C}$. Calculate the magnitude and direction of the net force on $q_C$.

  3. An isosceles triangle has two equal sides of $25\ \text{cm}$ and base $30\ \text{cm}$. Charges $q_1 = +5.0\ \mu\text{C}$ and $q_2 = +5.0\ \mu\text{C}$ are at the base corners, and $q_3 = -2.0\ \mu\text{C}$ is at the apex. Find the net force on $q_3$.

Score: ___/3

Part C: Electric Field Calculations (8 problems)

Target: 3-4 min per problem.

Set C1 — E-field from Point Charges (4 problems)

  1. A point charge $Q = +5.0\ \mu\text{C}$ is at the origin. Calculate the electric field at point $P$ located at $r = 40\ \text{cm}$ along the x-axis.

  2. Two point charges $q_1 = +6.0\ \mu\text{C}$ and $q_2 = -6.0\ \mu\text{C}$ are placed at $x = -20\ \text{cm}$ and $x = +20\ \text{cm}$ respectively (electric dipole). Calculate the electric field at: (a) The origin (b) Point $x = 40\ \text{cm}$ on the axis

  3. Three charges are arranged on the x-axis: $q_1 = +2.0\ \mu\text{C}$ at $x = 0$, $q_2 = -5.0\ \mu\text{C}$ at $x = 30\ \text{cm}$, and $q_3 = +3.0\ \mu\text{C}$ at $x = 60\ \text{cm}$. Find the net electric field at $x = 30\ \text{cm}$.

  4. Four charges are placed at the corners of a square with side $a = 40\ \text{cm}$: $q_1 = q_2 = q_3 = q_4 = +2.0\ \mu\text{C}$. Calculate the net electric field at the center of the square.

Score: ___/4

Set C2 — E-field Vector Components (4 problems)

  1. Two charges are on the x-axis: $Q_1 = +4.0\ \mu\text{C}$ at $x = 0$ and $Q_2 = -9.0\ \mu\text{C}$ at $x = 50\ \text{cm}$. Find the net electric field at point $P$ at $x = 20\ \text{cm}$ (magnitude and direction).

  2. For the configuration in problem 19, find where along the x-axis the electric field is zero.

  3. Three charges form a right triangle: $q_1 = +3.0\ \mu\text{C}$ at $(0, 0)$, $q_2 = -4.0\ \mu\text{C}$ at $(30\ \text{cm}, 0)$, and $q_3 = +5.0\ \mu\text{C}$ at $(0, 40\ \text{cm})$. Find the net electric field at the origin.

  4. Two charges $q_1 = +8.0\ \mu\text{C}$ and $q_2 = +2.0\ \mu\text{C}$ are separated by $d = 60\ \text{cm}$. Find the point(s) where the net electric field is zero, and determine if such a point exists between the charges or outside them.

Score: ___/4

Part D: Force on Charge in Electric Field (6 problems)

Target: 3 min per problem.

Set D1 — Force and Acceleration (3 problems)

  1. A charge $q = +3.0\ \mu\text{C}$ is placed in a uniform electric field $E = 500\ \text{N/C}$ directed east. Calculate: (a) The force on the charge (b) The acceleration if the charge has mass $m = 2.0 \times 10^{-4}\ \text{kg}$

  2. An electron is placed in a uniform electric field $E = 2000\ \text{N/C}$ directed upward. Calculate: (a) The force on the electron (b) The acceleration of the electron (c) The direction of acceleration

  3. A proton ($q = +e$, $m = 1.67 \times 10^{-27}\ \text{kg}$) is released from rest in a uniform electric field $E = 800\ \text{N/C}$. How far does it travel in $t = 2.0\ \mu\text{s}$?

Score: ___/3

Set D2 — Equilibrium and Suspended Charges (3 problems)

  1. A small sphere of mass $m = 0.50\ \text{g}$ carries a charge $q = +4.0\ \mu\text{C}$ and is suspended in a uniform electric field. What electric field strength is required to keep the sphere stationary?

  2. Two identical small spheres, each with mass $m = 2.0\ \text{g}$ and charge $q$, are suspended from a common point by light strings of length $L = 30\ \text{cm}$. The strings make an angle $\theta = 10°$ with the vertical. Find the charge $q$ on each sphere.

  3. A charged pith ball of mass $m = 0.20\ \text{g}$ is suspended by a thread in a horizontal electric field $E = 3000\ \text{N/C}$. If the thread makes an angle $\theta = 15°$ with the vertical, what is the charge on the ball?

Score: ___/3

Part E: Electron Motion in Uniform E-field (4 problems)

Target: 4-5 min per problem. (Most CQ1-style problems)

  1. An electron enters horizontally between two parallel plates with velocity $v_0 = 2.5 \times 10^{7}\ \text{m/s}$. The plates have length $L = 6.0\ \text{cm}$, separation $d = 2.0\ \text{cm}$, and potential difference $\Delta V = 200\ \text{V}$. Calculate: (a) The electric field between the plates (b) The vertical acceleration of the electron (c) The vertical deflection as it exits the plates (d) The exit angle $\theta$

  2. An electron with initial velocity $v_0 = 3.0 \times 10^{6}\ \text{m/s}$ enters a uniform electric field $E = 1500\ \text{N/C}$ directed vertically upward. The plates are $L = 5.0\ \text{cm}$ long. Calculate: (a) The time to traverse the plates (b) The vertical velocity component when exiting (c) The total deflection angle

  3. In a cathode ray tube, electrons are accelerated through a potential difference and then enter between deflection plates of length $L = 4.0\ \text{cm}$ with $E = 2500\ \text{N/C}$. If the deflection on the screen (located $D = 30\ \text{cm}$ beyond the plates) is $y = 2.4\ \text{cm}$, find the initial horizontal velocity of the electrons.

  4. A proton enters between parallel plates of length $L = 8.0\ \text{cm}$ and separation $d = 4.0\ \text{cm}$ with $v_0 = 5.0 \times 10^{5}\ \text{m/s}$. The plates are connected to $V = 500\ \text{V}$. Calculate: (a) Whether the proton will hit the upper plate before exiting (b) If not, the vertical deflection when it exits

Score: ___/4

Final Scorecard

Part Topic Problems Raw Score
A — 1D Coulomb Sets A1-A2 8 ___/8
B — 2D Coulomb Sets B1-B2 6 ___/6
C — E-field Sets C1-C2 8 ___/8
D — Force on q Sets D1-D2 6 ___/6
E — e⁻ Motion Set E 4 ___/4
TOTAL 32 ___/32

Proficiency Benchmarks for CQ1

Level Score Meaning
Exam-Ready $\ge 29/32$ (90%+) Ready for CQ1. Focus on speed.
Solid $25-28/32$ (78-87%) Good grasp. Review missed concepts.
Developing $20-24/32$ (62-75%) Drill weak areas again.
Needs Work $< 20/32$ (<62%) Re-study cheat sheet, retry tutorial.

Speed Benchmarks (CQ1: 15 min for 4 questions = 3.75 min/q)

  • < 90 min: Excellent speed for CQ1 conditions.
  • 90–110 min: Good. Practice under time pressure.
  • > 110 min: Need more pattern recognition drills.

Error Log Template

After grading, list every wrong problem with a one-word reason:

Problem Reason
e.g. 5 sign error

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Part A — Coulomb's Law 1D

  1. $F = k\frac{|q_1 q_2|}{r^2} = (9 \times 10^9)\frac{(3.0 \times 10^{-6})(5.0 \times 10^{-6})}{(0.20)^2} = 3.38\ \text{N}$ (attractive)

  2. $r = \sqrt{k\frac{e^2}{F}} = \sqrt{(9 \times 10^9)\frac{(1.6 \times 10^{-19})^2}{2.3 \times 10^{-8}}} = 1.0 \times 10^{-10}\ \text{m} = 1.0\ \text{Å}$

  3. $q = \sqrt{\frac{Fr^2}{k}} = \sqrt{\frac{(0.36)(0.15)^2}{9 \times 10^9}} = 3.0 \times 10^{-6}\ \text{C} = 3.0\ \mu\text{C}$

  4. Let $x$ be distance from $q_1$: $\frac{k(8)}{x^2} = \frac{k(2)}{(0.30-x)^2}$ → $x = 20\ \text{cm}$ from $q_1$ (between them)

  5. On $q_2$: $F_{12} = k\frac{(2)(4)}{(0.30)^2} = 0.80\ \text{N}$ (left), $F_{32} = k\frac{(6)(4)}{(0.30)^2} = 2.40\ \text{N}$ (left), $F_{\text{net}} = 3.20\ \text{N}$ left

  6. On $q_1$: $F_{21} = 0.80\ \text{N}$ (right), $F_{31} = k\frac{(6)(2)}{(0.60)^2} = 0.30\ \text{N}$ (right), $F_{\text{net}} = 1.10\ \text{N}$ right

  7. On $q_B$: $F_{AB} = k\frac{(5)(3)}{(0.10)^2} = 1.35 \times 10^{-5}\ \text{N}$ (right), $F_{CB} = k\frac{(4)(3)}{(0.15)^2} = 4.8 \times 10^{-6}\ \text{N}$ (left), $F_{\text{net}} = 8.7 \times 10^{-6}\ \text{N}$ right

  8. Zero force point must be outside, closer to smaller charge. Let $x$ be distance from $q_1$: $\frac{4}{x^2} = \frac{9}{(x+0.50)^2}$ → $x = 1.0\ \text{m}$ left of $q_1$

Part B — Coulomb's Law 2D

  1. $r_{12} = 0.30\ \text{m}$, $r_{13} = 0.40\ \text{m}$, $\vec{F}{21}$ along $-y$: $1.50\ \text{N}$, $\vec{F}{31}$ along $-x$: $0.844\ \text{N}$, $F_{\text{net}} = 1.72\ \text{N}$ at $\theta = 209°$

  2. $r_{13} = 0.60\ \text{m}$, $r_{23} = 1.0\ \text{m}$, $\vec{F}{13} = 0.25\ \text{N}$ at $53°$ above $+x$, $\vec{F}{23} = 0.216\ \text{N}$ at $37°$ above $-x$, $F_{\text{net}} = 0.32\ \text{N}$ at $\approx 7°$ above $+x$

  3. $F_{\text{net}} = \sqrt{F_{21}^2 + F_{31}^2} = \sqrt{(0.40)^2 + (0.225)^2} = 0.46\ \text{N}$ at $45°$ (along angle bisector)

  4. By symmetry, $F_{13} = F_{23} = k\frac{(4)(4)}{(0.20)^2} = 3.6\ \text{N}$. Angle between them = $60°$, so $F_{\text{net}} = 2(3.6)\cos(30°) = 6.24\ \text{N}$ upward

  5. $F_{AC} = F_{BC} = k\frac{(6)(3)}{(0.50)^2} = 0.648\ \text{N}$. Angle = $60°$, $F_{\text{net}} = 2(0.648)\cos(30°) = 1.12\ \text{N}$ directly toward midpoint of AB

  6. Height $h = \sqrt{25^2 - 15^2} = 20\ \text{cm}$. $F_{13} = F_{23} = k\frac{(5)(2)}{(0.25)^2} = 1.44\ \text{N}$. Vertical components add: $F_{\text{net}} = 2(1.44)(20/25) = 2.30\ \text{N}$ downward

Part C — Electric Field

  1. $E = k\frac{Q}{r^2} = (9 \times 10^9)\frac{5.0 \times 10^{-6}}{(0.40)^2} = 2.81 \times 10^{5}\ \text{N/C}$, radially outward

  2. (a) At origin: $E_1 = E_2 = k\frac{6 \times 10^{-6}}{(0.20)^2} = 1.35 \times 10^{6}\ \text{N/C}$, both point right, $E_{\text{net}} = 2.70 \times 10^{6}\ \text{N/C}$ right (b) At $x=40$: $E_1 = 3.75 \times 10^{5}\ \text{N/C}$ right, $E_2 = 1.50 \times 10^{6}\ \text{N/C}$ left, $E_{\text{net}} = 1.13 \times 10^{6}\ \text{N/C}$ left

  3. At $x=30$: $E_1$ (right) = $2.0 \times 10^{5}\ \text{N/C}$, $E_2$ undefined (at charge), $E_3$ (left) = $3.33 \times 10^{4}\ \text{N/C}$. Wait — need to recalculate. At $x=30$: $E_1 = k\frac{2}{(0.30)^2}$ right, $E_2$ diverges (on top of charge), $E_3 = k\frac{3}{(0.30)^2}$ left. If asking field at that point due to others: $E_{\text{net}} = k\frac{2}{(0.30)^2} - k\frac{3}{(0.30)^2}$ = negative = left

  4. By symmetry, fields from opposite corners cancel. $E_{\text{net}} = 0$

  5. $E_1 = k\frac{4}{(0.20)^2} = 9.0 \times 10^{5}\ \text{N/C}$ right, $E_2 = k\frac{9}{(0.30)^2} = 9.0 \times 10^{5}\ \text{N/C}$ right, $E_{\text{net}} = 1.80 \times 10^{6}\ \text{N/C}$ right

  6. Must be left of $q_1$ or right of $q_2$. Left of $q_1$: $\frac{4}{x^2} = \frac{9}{(x+0.50)^2}$ → $x = 1.0\ \text{m}$ left of $q_1$

  7. $E_2 = k\frac{4}{(0.30)^2} = 4.0 \times 10^{5}\ \text{N/C}$ left, $E_3 = k\frac{5}{(0.40)^2} = 2.81 \times 10^{5}\ \text{N/C}$ down, $E_{\text{net}} = 4.89 \times 10^{5}\ \text{N/C}$ at $215°$

  8. Cannot be between (both fields point same direction). Left of $q_2$: $\frac{8}{(0.60+x)^2} = \frac{2}{x^2}$ → $x = 0.60\ \text{m}$ left of $q_2$ (or at $x = -0.60\ \text{m}$ from origin if $q_2$ at origin)

Part D — Force on Charge

  1. (a) $F = qE = (3.0 \times 10^{-6})(500) = 1.5 \times 10^{-3}\ \text{N} = 1.5\ \text{mN}$ east (b) $a = F/m = (1.5 \times 10^{-3})/(2.0 \times 10^{-4}) = 7.5\ \text{m/s}^2$ east

  2. (a) $F = eE = (1.6 \times 10^{-19})(2000) = 3.2 \times 10^{-16}\ \text{N}$ downward (b) $a = F/m_e = (3.2 \times 10^{-16})/(9.11 \times 10^{-31}) = 3.51 \times 10^{14}\ \text{m/s}^2$ (c) Downward (opposite to E-field for negative charge)

  3. $a = eE/m_p = (1.6 \times 10^{-19})(800)/(1.67 \times 10^{-27}) = 7.67 \times 10^{10}\ \text{m/s}^2$, $d = \frac{1}{2}at^2 = 0.5(7.67 \times 10^{10})(2.0 \times 10^{-6})^2 = 0.153\ \text{m} = 15.3\ \text{cm}$

  4. $qE = mg$ → $E = mg/q = (5.0 \times 10^{-4})(9.8)/(4.0 \times 10^{-6}) = 1.23 \times 10^{3}\ \text{N/C}$ upward

  5. $T\cos\theta = mg$, $T\sin\theta = F_e = k\frac{q^2}{(2L\sin\theta)^2}$. Solve: $q = 2L\sin\theta\sqrt{\frac{mg\tan\theta}{k}} = 1.56 \times 10^{-7}\ \text{C} = 0.156\ \mu\text{C}$

  6. $qE = mg\tan\theta$ → $q = mg\tan\theta/E = (2.0 \times 10^{-4})(9.8)\tan(15°)/(3000) = 1.75 \times 10^{-7}\ \text{C} = 0.175\ \mu\text{C}$

Part E — Electron Motion

  1. (a) $E = V/d = 200/0.02 = 1.0 \times 10^{4}\ \text{N/C}$ (b) $a = eE/m_e = (1.6 \times 10^{-19})(10^4)/(9.11 \times 10^{-31}) = 1.76 \times 10^{15}\ \text{m/s}^2$ upward (c) $t = L/v_0 = 0.06/(2.5 \times 10^7) = 2.4 \times 10^{-9}\ \text{s}$, $y = \frac{1}{2}at^2 = 0.5(1.76 \times 10^{15})(2.4 \times 10^{-9})^2 = 5.07 \times 10^{-3}\ \text{m} = 5.07\ \text{mm}$ (d) $v_y = at = 4.22 \times 10^{6}\ \text{m/s}$, $\theta = \tan^{-1}(v_y/v_0) = 9.63°$

  2. (a) $t = L/v_0 = 0.05/(3.0 \times 10^6) = 1.67 \times 10^{-8}\ \text{s}$ (b) $a = eE/m_e = 2.64 \times 10^{14}\ \text{m/s}^2$ downward, $v_y = at = 4.40 \times 10^{6}\ \text{m/s}$ downward (c) $\theta = \tan^{-1}(4.40/30) = 8.35°$ below horizontal

  3. $y_{\text{plates}} = \frac{1}{2}\frac{eE}{m}\left(\frac{L}{v_0}\right)^2$, $v_y = \frac{eEL}{mv_0}$, additional deflection $y_{\text{after}} = v_y \cdot \frac{D}{v_0}$. Total: solve for $v_0 = 2.0 \times 10^{7}\ \text{m/s}$

  4. (a) $E = V/d = 12500\ \text{N/C}$, $a = eE/m_p = 1.20 \times 10^{12}\ \text{m/s}^2$ downward, $t = L/v_0 = 1.6 \times 10^{-7}\ \text{s}$, $y_{\text{max}} = 0.5(1.20 \times 10^{12})(1.6 \times 10^{-7})^2 = 1.54 \times 10^{-2}\ \text{m} = 1.54\ \text{cm} < 2.0\ \text{cm}$, so doesn't hit (b) $y = 1.54\ \text{cm}$

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