⏱️ Duration: 75 minutes | Total: /60 marks. Show all work, including diagrams. Use \(k=8.99\times10^9\,\text{N·m}^2/\text{C}^2\), \(G=6.67\times10^{-11}\,\text{N·m}^2/\text{kg}^2\), \(e=1.60\times10^{-19}\,\text{C}\), \(M_E=5.98\times10^{24}\,\text{kg}\), \(R_E=6.38\times10^{6}\,\text{m}\).
Two point charges \(q_1=+5.0\,\mu\text{C}\) and \(q_2=-3.0\,\mu\text{C}\) are separated by \(0.20\,\text{m}\). Calculate the magnitude and state the direction of the electric force on \(q_1\).
2. [3]
A satellite orbits Earth at an altitude of \(400\,\text{km}\). Calculate (a) the gravitational field strength at this altitude, (b) the orbital speed, (c) the orbital period.
3. [2]
Sketch the electric field pattern between two oppositely charged parallel plates. Mark the direction of the field, indicate where the field is uniform, and label the high-V and low-V plates.
[ Sketch parallel plates with field lines from + to − plate ]
4. [3]
An electron (\(m=9.11\times10^{-31}\,\text{kg}\)) is accelerated through a potential difference of \(2500\,\text{V}\). Calculate its final speed using energy conservation.
5. [2]
A current of \(4.0\,\text{A}\) flows in a wire of length \(0.50\,\text{m}\) placed perpendicular to a magnetic field of \(0.30\,\text{T}\). Calculate the magnetic force on the wire.
6. [3]
State, with brief justification, whether each is true or false:
(a) Electric field lines can cross.
(b) The electric field inside a conductor in equilibrium is zero.
(c) Magnetic forces can do work on a moving charged particle.
Part B: Thinking & Investigation [15 marks]
7. [5]
Three charges are placed at corners of an equilateral triangle of side \(0.10\,\text{m}\): \(q_A=+2.0\,\mu\text{C}\), \(q_B=+2.0\,\mu\text{C}\), \(q_C=-2.0\,\mu\text{C}\). Calculate the magnitude and direction of the net electric force on \(q_C\). Show the vector diagram.
[ Draw triangle with charges and force vectors on q_C ]
8. [5]
An alpha particle (\(q=+2e\), \(m=6.64\times10^{-27}\,\text{kg}\)) enters a uniform magnetic field of \(0.40\,\text{T}\) at a speed of \(2.0\times10^{6}\,\text{m/s}\) perpendicular to the field. (a) Show that the path is a circle. (b) Calculate the radius. (c) Calculate the period of the motion.
9. [5]
In Millikan's oil drop experiment, a charged droplet of mass \(3.2\times10^{-15}\,\text{kg}\) is held stationary between parallel plates separated by \(1.5\,\text{cm}\) when the potential difference is \(490\,\text{V}\). (a) Determine the charge on the droplet. (b) How many excess electrons does it carry? (c) Justify the conclusion that charge is quantized.
Part C: Communication [15 marks]
10. [5]
Compare and contrast gravitational and electric fields. Address: source of the field, sign of "charge", magnitude of \(k\) vs. \(G\), shielding, and one specific phenomenon that depends on each.
11. [5]
A student claims: "A magnetic field exerts a force on a charged particle whether or not it moves." Correct this misconception with clear reasoning, including the formula \(F = qvB\sin\theta\) and an example.
12. [5]
Using a labelled diagram, explain how a mass spectrometer separates ions of different mass-to-charge ratio. Include the velocity selector (E and B), the deflection chamber (B only), and derive the relationship \(m/q = Br/v\).
[ Mass spectrometer: velocity selector → deflection region ]
Part D: Application [15 marks]
13. [5]
A geostationary communications satellite orbits Earth with a period of \(24\,\text{h}\). (a) Calculate the orbital radius. (b) Calculate the orbital speed. (c) Discuss one technological benefit and one social/environmental concern of geostationary satellites.
14. [5]
An inkjet printer deflects charged ink droplets between parallel plates. A droplet of mass \(2.0\times10^{-10}\,\text{kg}\) and charge \(-3.0\times10^{-13}\,\text{C}\) enters horizontally at \(15\,\text{m/s}\) midway between plates that are \(0.5\,\text{cm}\) long with field \(2.0\times10^{6}\,\text{N/C}\) directed downward. Find the vertical deflection on exit.
15. [5]
A simple DC motor uses a rectangular loop \((L=8.0\,\text{cm}, W=5.0\,\text{cm})\) of \(50\) turns carrying \(2.5\,\text{A}\) in a \(0.10\,\text{T}\) field. (a) Calculate the maximum torque on the loop. (b) Explain why the torque varies with orientation. (c) Discuss two STSE implications of replacing internal-combustion engines with electric motors.
(a) F — uniqueness of field direction at each point. (b) T — free charges redistribute to cancel internal field. (c) F — magnetic force is always perpendicular to \(\vec v\), hence \(\vec F\cdot\vec v=0\).
Q7.
\(F_{AC}=F_{BC}=kq^2/r^2=8.99e9(2e{-}6)^2/0.01=3.60\,\text{N}\). By symmetry the components along AB cancel; net force points away from midpoint of AB along the perpendicular: \(F_{net} = 2F\cos 30° = 6.23\,\text{N}\) (attractive, so toward AB midpoint since C is negative). Direction: along the altitude from C toward midpoint of AB.
Q8.
(a) \(\vec F\) is always \(\perp \vec v\) and constant in magnitude → uniform circular motion. (b) \(r = mv/(qB) = (6.64e{-}27)(2e6)/((3.2e{-}19)(0.40))=0.104\,\text{m}\). (c) \(T=2\pi m/(qB)=3.26\times10^{-7}\,\text{s}\).
Both inverse-square central fields; gravity always attractive (mass only +), electric attractive or repulsive. \(k\gg G\); electric force can be screened by conductors, gravity cannot. Phenomena: orbits (gravity); lightning (electric).
Q11.
Magnetic force = \(qvB\sin\theta\). For stationary charge \(v=0\) → \(F=0\). Example: a stationary electron near a bar magnet experiences no magnetic force, but a moving electron does.
Q12.
Velocity selector: \(qE = qvB_1 \Rightarrow v=E/B_1\). In deflection region: \(qvB_2 = mv^2/r \Rightarrow m/q = B_2 r/v\). Different masses → different radii → ions strike detector at different positions.
Q13.
(a) \(T^2=4\pi^2 r^3/(GM_E)\Rightarrow r=\sqrt[3]{GM_E T^2/(4\pi^2)}=4.22\times10^7\,\text{m}\) (\(\approx 35{,}900\,\text{km}\) altitude). (b) \(v=2\pi r/T=3.07\times10^3\,\text{m/s}\). (c) Benefit: continuous coverage for telecom/weather. Concern: orbital debris and electronic-waste from end-of-life satellites.
Q14.
\(t = L/v = 0.005/15 = 3.33\times10^{-4}\,\text{s}\). \(F = qE = 6.0\times10^{-7}\,\text{N}\) (upward since charge is negative and field is down). \(a = F/m = 3000\,\text{m/s}^2\). \(\Delta y = \tfrac12 at^2 = 1.66\times10^{-4}\,\text{m} \approx 0.17\,\text{mm}\) upward.
Q15.
(a) \(\tau_{max} = NBIA = 50(0.10)(2.5)(0.08\cdot0.05) = 5.0\times10^{-2}\,\text{N·m}\). (b) \(\tau = NBIA\sin\phi\); zero when plane of coil is parallel to B. (c) Reduced tailpipe emissions (benefit); increased grid demand and battery raw-material extraction (concern).