Electrostatics in NET covers Coulomb's law, electric field and potential, Gauss's law and capacitors. The numericals are short, but the powers of ten are not, so most lost marks come from μC, cm and k = 9 × 10⁹ handled carelessly.
Key facts
- Physics is 30% of NET-Engineering, a derived 60 of the 200 MCQs.
- Electrostatics appears twice in the syllabus: in first-year Electrostatics and Current Electricity, and again as second-year Electrostatics with Gauss's law.
- Our syllabus data estimates about 6 questions from the second-year chapter, and about 8 from the first-year chapter shared with current electricity.
Source: NUST NET-Engineering composition and the SailToNUST physics syllabus, checked October 2026
What the chapter covers
The second-year syllabus line is short: Coulomb's law, electric field and potential, capacitance, and Gauss's law. The first-year chapter covers the same Coulomb and capacitor material, then moves into circuits. This post takes the electrostatics half. Circuits, resistance and power are in the current electricity guide.
Formulas you need
Coulomb's law F = k q₁q₂ / r²
k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻²
ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²
In a dielectric F = k q₁q₂ / (εᵣ r²)
Electric field E = F/q = kq/r² N C⁻¹ (same as V m⁻¹)
Between parallel plates E = V/d
Potential of a point charge V = kq/r V = J C⁻¹
Work moving a charge W = qΔV
Electron-volt 1 eV = 1.6 × 10⁻¹⁹ J
Electric flux Φ = EA cosθ N m² C⁻¹
Gauss's law Φ = Q_enclosed / ε₀
Capacitance C = Q/V F = C V⁻¹
Parallel plates C = ε₀εᵣA / d
Energy stored U = ½CV² = ½QV = Q²/2C
Series 1/C = 1/C₁ + 1/C₂ + ...
Parallel C = C₁ + C₂ + ...
RC time constant τ = RC s
Worked MCQs
1. Two charges in air. Charges of 3 μC and 4 μC sit 30 cm apart in air. What is the force between them?
(a) 0.12 N (b) 1.2 N (c) 12 N (d) 0.4 N
F = 9 × 10⁹ × (3 × 10⁻⁶)(4 × 10⁻⁶) / (0.3)²
= 9 × 10⁹ × 12 × 10⁻¹² / 0.09
= 0.108 / 0.09 = 1.2 N
Answer (b). Put the same pair in oil with εᵣ = 4 and the force drops to 0.3 N.
2. Touch and separate. Two identical small conducting spheres carry +10 μC and −20 μC and attract with force F₁ at distance R. They are touched together and returned to distance R, where the force is F₂. What is F₁ : F₂?
(a) 1 : 8 (b) −8 : 1 (c) 1 : 2 (d) −2 : 1
After contact, total charge −10 μC splits equally: −5 μC each
F₁ ∝ (+10)(−20) = −200 (attraction)
F₂ ∝ (−5)(−5) = +25 (repulsion)
F₁ : F₂ = −200 : 25 = −8 : 1
Answer (b). The minus sign only records that one force attracts and the other repels. Identical spheres share charge equally. Spheres of different radius do not, so look for the word "identical" before you split evenly.
3. Capacitors in series. A 3 μF and a 6 μF capacitor are joined in series across 12 V. What is the voltage across the 3 μF capacitor?
(a) 4 V (b) 6 V (c) 8 V (d) 12 V
C = (3 × 6)/(3 + 6) = 2 μF
Q = CV = 2 × 12 = 24 μC (same on both in series)
V₃ = Q/C = 24/3 = 8 V, V₆ = 24/6 = 4 V
Answer (c). In series the smaller capacitor takes the larger share of the voltage. That is the reverse of resistors in series, and it is why (a) is a tempting wrong answer.
4. Energy and time constant. A 2 μF capacitor charges to 100 V through a 1 MΩ resistor. How much energy does it store, and what is the time constant?
U = ½CV² = ½ × 2 × 10⁻⁶ × 100² = 0.01 J
τ = RC = 10⁶ × 2 × 10⁻⁶ = 2 s
After one time constant the capacitor holds 63% of its final charge. It never reaches 100% in theory, though it gets past 99% after about 5τ.
5. Electron through a potential difference. An electron starts from rest and accelerates through 1000 V. What is its final speed? (mₑ = 9.11 × 10⁻³¹ kg)
(a) 1.9 × 10⁵ m/s (b) 1.9 × 10⁷ m/s (c) 3.5 × 10¹⁴ m/s (d) 6 × 10⁸ m/s
K = eV = 1.6 × 10⁻¹⁹ × 1000 = 1.6 × 10⁻¹⁶ J (= 1000 eV)
v = √(2K/m) = √(3.2 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = √(3.51 × 10¹⁴)
v ≈ 1.87 × 10⁷ m/s
Answer (b). Option (c) is v² with the square root forgotten. Option (d) is faster than light, which should rule it out on sight.
Traps that cost marks
- Newton's third law applies. A 1 μC and a 5 μC charge feel forces of equal size. The ratio is 1 : 1, whatever the charges are.
- The field inside a conductor is zero. All the charge sits on the outer surface. The potential inside is constant and equal to the surface value, not zero.
- Work around a closed path is zero. The electrostatic force is conservative. Moving a charge round any loop, or along an equipotential, takes no net work.
- Flux depends only on enclosed charge. The size or shape of the closed surface does not matter. A charge q at the centre of a cube sends q/6ε₀ through each face.
- Dielectric with battery attached or not. Insert a dielectric with the battery connected and V stays fixed, so Q and U go up. Insert it after disconnecting and Q stays fixed, so V and U go down. Read which case the question describes.
- Force is largest in air or vacuum. Any material medium has εᵣ greater than 1 and weakens it.
Speed tips for the exam
Work in powers of ten separately. For Coulomb questions, multiply the mantissas (9 × 3 × 4 = 108) and add the exponents (9 − 6 − 6 = −3) in one pass, then divide by r². That gets you 0.108 / 0.09 in your head.
Memorise that 1 cm² is 10⁻⁴ m² and 1 mm² is 10⁻⁶ m². Parallel-plate capacitance questions almost always give area in cm².
For series capacitors, two equal ones give half. For two unequal ones, use product over sum, the same shortcut you use for parallel resistors.
FAQ
Is Gauss's law in the NET syllabus?
Yes. It is listed in the second-year Electrostatics chapter. Expect concept questions on flux and enclosed charge more than heavy integration.
What is the value of k in Coulomb's law?
k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻². Use it unless the question gives ε₀ and expects you to work with that instead.
Why are electric field lines never closed loops?
Electrostatic field lines start on positive charges and end on negative ones. A closed loop would mean non-zero work round a closed path, which a conservative field cannot do.
Which physics formulas should I memorise for NET?
For this chapter, the boxed list above. For every chapter on one page, use the physics formula sheet.
