This checkpoint covers Lessons 1 to 4: electric charge, charging processes, Coulomb's Law, electric fields, field lines, potential difference, work done, and charged particle motion in uniform fields.
Checkpoint Assessment
1. Two identical conducting spheres carry charges of +6 C and −2 C. They are touched together and then separated. What is the final charge on each sphere?
2. The electrostatic force between two point charges is $F$ when they are distance $r$ apart. If the distance is halved to $r/2$, what is the new force?
3. Which statement correctly describes the electric field lines around a negative point charge?
4. A neutral conductor is charged by induction using a positively charged rod. What is the sign of the final charge on the conductor?
5. Two parallel plates are separated by 2.0 cm and connected to a 300 V power supply. What is the electric field strength between the plates?
6. Which of the following is a correct similarity between Coulomb's Law and Newton's Law of Gravitation?
7. An electron is accelerated from rest through a potential difference of 100 V. Which expression gives the work done on the electron?
8. An electron enters a uniform electric field perpendicular to its initial velocity. What is the shape of its trajectory?
9. A student rubs a plastic ruler with a woollen cloth and the ruler becomes negatively charged. Explain this process in terms of electron transfer and conservation of charge. 3 MARKS
10. Two point charges, $q_1 = +3.0 \times 10^{-8}\ \text{C}$ and $q_2 = -5.0 \times 10^{-8}\ \text{C}$, are placed 20 cm apart in air. Calculate the magnitude of the electrostatic force between them and state whether it is attractive or repulsive. 3 MARKS
11. An electron is fired horizontally at $4.0 \times 10^6\ \text{m s}^{-1}$ into the space between two horizontal parallel plates of length 2.0 cm and separation 1.0 cm. The plates are connected to a 200 V power supply. Calculate the vertical deflection of the electron as it exits the plates. 4 MARKS
1. C — Total charge = +6 + (−2) = +4 C. Each identical sphere gets half: +2 C.
2. B — Coulomb's Law is inverse-square: $F \propto 1/r^2$. Halving $r$ increases $F$ by factor of 4.
3. D — Field lines point inward toward a negative charge (the direction a positive test charge would be pulled).
4. A — Induction with a positive rod repels electrons to the far side; grounding allows electrons to escape, leaving the conductor negatively charged.
5. C — $E = V/d = 300/(2.0 \times 10^{-2}) = 1.5 \times 10^4\ \text{V m}^{-1}$.
6. B — Both follow inverse-square laws. Coulomb's force can be repulsive; gravity is always attractive.
7. D — $W = qV = (1.60 \times 10^{-19})(100) = 1.60 \times 10^{-17}\ \text{J}$. The magnitude of charge is used for work calculation.
8. A — Constant horizontal velocity and constant vertical acceleration produce a parabolic trajectory (projectile motion analogy).
Q9 (3 marks): When the plastic ruler is rubbed with wool, friction causes electrons to transfer from the wool to the ruler. The ruler gains excess electrons and becomes negatively charged. By the law of conservation of charge, the wool loses the same number of electrons and becomes positively charged with equal magnitude. The total charge of the ruler-wool system remains zero throughout the process.
Q10 (3 marks): Using Coulomb's Law: $F = k|q_1 q_2|/r^2 = (8.99 \times 10^9)(3.0 \times 10^{-8})(5.0 \times 10^{-8})/(0.20)^2 = (8.99 \times 10^9)(1.5 \times 10^{-15})/0.04 = 3.37 \times 10^{-4}\ \text{N}$. The force is attractive because the charges are opposite ($+/-$).
Q11 (4 marks): $E = V/d = 200/(1.0 \times 10^{-2}) = 2.0 \times 10^4\ \text{V m}^{-1}$. $a = |q_e|E/m_e = (1.60 \times 10^{-19})(2.0 \times 10^4)/(9.11 \times 10^{-31}) = 3.51 \times 10^{15}\ \text{m s}^{-2}$. $t = L/u = (2.0 \times 10^{-2})/(4.0 \times 10^6) = 5.0 \times 10^{-9}\ \text{s}$. $y = \tfrac{1}{2}at^2 = 0.5(3.51 \times 10^{15})(5.0 \times 10^{-9})^2 = 4.39 \times 10^{-2}\ \text{m} = 4.4\ \text{cm}$.
Tick when you have finished the checkpoint and checked the answers.