Electrostatics MCQs
MDCAT · Physics · 6 questions
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Revision summary
PMDC MDCAT syllabus: Electrostatics
Coulomb's law: the force between two point charges is F = kq₁q₂/r², with k ≈ 9 × 10⁹ N m² C⁻²; it follows an inverse square law, so doubling the distance quarters the force. Charge is quantised in units of e = 1.6 × 10⁻¹⁹ C. Electric field intensity E = F/q is measured in N/C (equal to V/m), and field lines run from positive to negative charges, never cross, and crowd together where the field is strong. Electric potential is work per unit charge, in volts (J/C). A capacitor stores charge: C = Q/V, in farads. For a parallel-plate capacitor, C = ε₀A/d, and a dielectric multiplies it by εᵣ. Capacitors add directly in parallel (C = C₁ + C₂) and as reciprocals in series (1/C = 1/C₁ + 1/C₂), the reverse of resistors. Energy stored is ½CV². Typical MDCAT items test the inverse square law, units and series/parallel combinations.
Questions with answers
Written by Parcha from the syllabus. Samples until a subject reviewer checks them; report anything wrong from the practice screen.
Q1 · Parcha original · Sample
If the distance between two point charges is doubled, the electric force between them becomes:
- A. Half
- B. One quarter
- C. Double
- D. Four times
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Answer: B. One quarter
Coulomb's law: F = kq₁q₂/r². Doubling r multiplies r² by 4, so the force falls to one quarter.
Open this question on its own pageQ2 · Parcha original · Sample
The magnitude of the charge on an electron is:
- A. 1.6 × 10⁻¹⁹ C
- B. 9.1 × 10⁻³¹ C
- C. 1.6 × 10¹⁹ C
- D. 6.25 × 10¹⁸ C
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Answer: A. 1.6 × 10⁻¹⁹ C
e = 1.6 × 10⁻¹⁹ C (the electron's charge is −e). 9.1 × 10⁻³¹ kg is the electron's mass, and 6.25 × 10¹⁸ is the number of electrons in one coulomb.
Q3 · Parcha original · Sample
The SI unit of electric field intensity is:
- A. N C
- B. C/N
- C. J/C
- D. N/C
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Answer: D. N/C
Field intensity E = F/q, so its unit is newton per coulomb (N/C), which equals volt per metre (V/m). J/C is the volt, the unit of potential.
Q4 · Parcha original · Sample
The SI unit of capacitance is the:
- A. Henry
- B. Farad
- C. Tesla
- D. Ohm
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Answer: B. Farad
Capacitance C = Q/V, measured in farads (1 F = 1 C/V). The henry is inductance, the tesla magnetic field, and the ohm resistance.
Q5 · Parcha original · Sample
Two 4 μF capacitors are connected in parallel. Their combined capacitance is:
- A. 2 μF
- B. 4 μF
- C. 8 μF
- D. 16 μF
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Answer: C. 8 μF
In parallel, capacitances add: 4 + 4 = 8 μF. In series they would give 1/C = 1/4 + 1/4, so C = 2 μF.
Q6 · Parcha original · Sample
Electric field lines:
- A. Start on negative and end on positive charges
- B. Form closed loops
- C. Never cross each other
- D. Are densest where the field is weakest
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Answer: C. Never cross each other
Field lines start on positive and end on negative charges, never cross (the field has one direction at each point), and are closest together where the field is strongest. Closed loops are a feature of magnetic field lines.