A point charge lies outside a closed Gaussian surface. The net electric flux through the surface is:

It depends on the distance of the charge

The electric field due to an infinite charged sheet of surface density σ:

Is σ/2ε₀, independent of distance

A charged capacitor of capacitance C at potential V is connected to an identical uncharged capacitor. The total stored energy becomes:

Unchanged, since charge is conserved

One quarter of its original value

Inside a hollow charged conducting shell, in electrostatic equilibrium:

E is zero and V is constant, equal to the surface value

Two capacitors of 2 μF and 3 μF are connected in series across a battery. Which is the same for both?

The potential difference across each

Electrostatics — Fields, Gauss's Law, Potential and Capacitors

Electrostatics is the largest single block in the physics syllabus and it rewards a specific habit: decide whether the question wants a vector or a scalar before you start. Field and force are vectors and must be added component by component; potential and energy are scalars and simply add with signs. Students who lose marks here almost always lost them by adding field magnitudes or by carrying a potential as though it had a direction. The chapter also introduces Gauss's law, which is the one tool in the syllabus that turns an intractable integral into a one-line answer whenever the charge distribution is symmetric.

Coulomb's Law, Field and the Vector Habit

Two point charges interact with F = kq₁q₂/r², where k = 1/4πε₀ ≈ 9 × 10⁹ N·m²/C². The field is the force per unit charge, E = F/q₀ = kq/r², pointing away from a positive charge and towards a negative one. In a medium of dielectric constant K the force is divided by K, which is why immersing charges in water weakens their interaction eighty-fold.

Vector or scalar — decide first
Quantity	Nature	How to combine several sources
Force F, field E	Vector	Resolve into components and add those; never add magnitudes
Potential V, energy U	Scalar	Add algebraically, keeping the sign of each charge
Flux Φ	Scalar	Φ = ∮E⃗·dA⃗, so orientation matters through the dot product

⚠️ E can be zero where V is not, and V can be zero where E is not

These two are independent and questions exploit that constantly. Midway between two equal positive charges, the fields cancel by symmetry so E = 0, but both potentials are positive so V ≠ 0. Midway between equal and opposite charges, the potentials cancel so V = 0, but the fields both point the same way and add, so E ≠ 0. Inside a charged conductor, E = 0 everywhere but V is constant and non-zero. So never infer one from the other. The relation between them is E = −dV/dr, which says that E is the rate of change of V — so a region of constant non-zero potential has zero field, exactly as a flat hill has no slope.

Standard fields worth knowing without derivation
Source	Field
Point charge	E = kq/r², so E ∝ 1/r²
Infinite line charge	E = λ/2πε₀r, so E ∝ 1/r
Infinite sheet of charge	E = σ/2ε₀ — independent of distance
Charged conducting shell, outside	E = kQ/r², as if all the charge were at the centre
Charged conducting shell, inside	E = 0, with V constant and equal to the surface value
Dipole, on the axis	E = 2kp/r³ — note the 1/r³

Gauss's Law — Symmetry Does the Integral for You

Gauss's law states that the net flux through any closed surface is Φ = q_enclosed/ε₀. Two features make it powerful: the flux depends only on the charge inside, and the surface can be any shape you like — so you choose one that makes E constant and perpendicular over it, and the integral collapses to E × Area.

🎯 Why a charge outside the surface contributes nothing

An external charge's field lines enter the closed surface and then leave it again, so every line contributes one negative unit of flux on the way in and one positive on the way out, cancelling exactly. That is why moving an external charge closer changes the field at every point on the surface yet leaves the total flux unchanged — a result that feels wrong and is asked precisely for that reason. Two consequences to carry: the flux is unchanged if you move a charge around inside the surface, since only the enclosed amount matters; and a surface enclosing zero net charge has zero flux even if it is sitting in a strong field.

🧠 The three symmetries, and which surface to choose

Gauss's law is only useful when the symmetry lets you pull E out of the integral, and there are exactly three cases in the syllabus. Spherical charge (point, ball, shell) → choose a concentric sphere, giving E(4πr²) = q/ε₀. Cylindrical charge (long wire, long cylinder) → choose a coaxial cylinder, giving E(2πrL) = λL/ε₀. Planar charge (large sheet) → choose a pillbox straddling the sheet, giving E(2A) = σA/ε₀. Learn the three surfaces rather than the three answers, because then a hollow sphere with a cavity, or a wire inside a cylindrical shell, is the same problem with a different q_enclosed.

Capacitors, Dielectrics and Energy

Capacitance is charge per unit potential difference, C = Q/V, and for a parallel-plate capacitor C = ε₀A/d, or Kε₀A/d with a dielectric filling the gap. The stored energy can be written three equivalent ways: U = ½QV = ½CV² = Q²/2C, and choosing the right one is most of the skill.

Series and parallel — the reverse of resistors
Arrangement	Capacitance	What is shared
Series	1/C = 1/C₁ + 1/C₂ — smaller than either	Charge is the same on each; voltages add
Parallel	C = C₁ + C₂ — larger than either	Voltage is the same across each; charges add

⚠️ Inserting a dielectric: battery connected or disconnected?

This single distinction generates a large family of questions, and the answers are opposite in the two cases. Battery still connected → V is fixed by the battery. C rises to KC, so Q = CV rises to KQ and the energy U = ½CV² rises to KU, with the extra energy coming from the battery. Battery disconnected → Q is fixed, because the charge has nowhere to go. C still rises to KC, so V = Q/C falls to V/K and the energy U = Q²/2C falls to U/K, the difference being the work done by the field pulling the dielectric in. So the same physical act raises the energy in one case and lowers it in the other. Always ask first which quantity is held fixed, then choose the energy form that contains it — ½CV² when V is fixed, Q²/2C when Q is fixed.

🎯 Why energy is lost when two capacitors are connected together

Connect a charged capacitor to an identical uncharged one. Charge conservation shares the charge equally, so each ends at V/2 — and the total energy, which was ½CV², becomes 2 × ½C(V/2)² = ¼CV², exactly half of what you started with. Where did it go? Into resistive heating and electromagnetic radiation in the connecting wires during the transient, however short the wires. This is a favourite question because it looks like a violation: charge is conserved but energy is not, and the resolution is that the process is irreversible. The general result is worth holding: connecting capacitors at different potentials always loses energy, and the loss is C₁C₂(V₁ − V₂)²/2(C₁ + C₂).

ℹ️ Conductors: three facts that solve most questions

In electrostatic equilibrium, the field inside a conductor is zero, all excess charge sits on the outer surface, and the whole conductor is at one potential. The first is why a metal box shields its contents — a Faraday cage, and why you are safe in a car during a lightning strike. The second is why charge density is highest where the surface curves most sharply, so a lightning conductor is pointed. The third means no work is needed to move a charge anywhere within or on a conductor. One more result: the field just outside the surface of a conductor is σ/ε₀ and always perpendicular to it, because any tangential component would immediately drive a surface current and so cannot persist.

Key takeaways

F = kq₁q₂/r² with k ≈ 9 × 10⁹; a dielectric of constant K divides the force by K.
E and F are vectors (add components); V and U are scalars (add with signs). Decide which before starting.
E = 0 does not imply V = 0, and V = 0 does not imply E = 0 — E is the rate of change of V.
Field of a point charge ∝ 1/r², a line ∝ 1/r, a sheet is independent of distance, a dipole axis ∝ 1/r³.
Gauss: Φ = q_enclosed/ε₀. External charges contribute zero net flux, and moving a charge inside changes nothing.
Capacitors: series adds reciprocals and shares charge; parallel adds directly and shares voltage.
Dielectric with the battery ON raises U to KU; with the battery OFF it lowers U to U/K.
Connecting two capacitors at different potentials always dissipates energy, though charge is conserved.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

At the midpoint between two equal positive point charges:
1. Both E and V are zero
2. E is zero but V is not
3. V is zero but E is not
4. Neither is zero
Show answer
Answer: B — E is zero but V is not
The two fields are equal in magnitude and opposite in direction, so they cancel and E = 0. Potential is a scalar and both contributions are positive, so they add and V ≠ 0. For equal and opposite charges the situation reverses — V = 0 while E ≠ 0 — which is why one can never be inferred from the other.
A point charge lies outside a closed Gaussian surface. The net electric flux through the surface is:
1. q/ε₀
2. Zero
3. q/2ε₀
4. It depends on the distance of the charge
Show answer
Answer: B — Zero
Every field line from an external charge that enters the surface also leaves it, so the inward and outward contributions cancel exactly and the net flux is zero. Moving the charge closer changes E at every point on the surface yet leaves the total flux unchanged, since Gauss's law depends only on the enclosed charge.
A capacitor is charged by a battery, the battery is DISCONNECTED, and then a dielectric of constant K is inserted. The stored energy:
1. Becomes K times larger
2. Becomes K times smaller
3. Is unchanged
4. Becomes K² times larger
Show answer
Answer: B — Becomes K times smaller
With the battery disconnected the charge Q is fixed, so use U = Q²/2C. Since C rises to KC, U falls to U/K, the difference being work done by the field as it pulls the dielectric in. Had the battery stayed connected, V would be fixed instead and U = ½CV² would rise to KU — the same physical act with the opposite answer.
The electric field due to an infinite charged sheet of surface density σ:
1. Varies as 1/r²
2. Varies as 1/r
3. Is σ/2ε₀, independent of distance
4. Is zero everywhere
Show answer
Answer: C — Is σ/2ε₀, independent of distance
Applying Gauss's law with a pillbox straddling the sheet gives E(2A) = σA/ε₀, so E = σ/2ε₀ with no r in it — moving away from an infinite sheet does not weaken its field, because the extra distance is exactly offset by the greater area of sheet that becomes relevant. The 1/r² and 1/r forms belong to a point charge and a line charge respectively.
A charged capacitor of capacitance C at potential V is connected to an identical uncharged capacitor. The total stored energy becomes:
1. Unchanged, since charge is conserved
2. Half of its original value
3. Twice its original value
4. One quarter of its original value
Show answer
Answer: B — Half of its original value
The charge divides equally so each capacitor sits at V/2, giving total energy 2 × ½C(V/2)² = ¼CV² against the original ½CV² — exactly half. The missing half is dissipated as heat and radiation in the connecting wires during the transient, however short they are. Charge is conserved; energy is not, because the process is irreversible.
Inside a hollow charged conducting shell, in electrostatic equilibrium:
1. Both E and V are zero
2. E is zero and V is constant, equal to the surface value
3. E is constant and non-zero
4. V decreases towards the centre
Show answer
Answer: B — E is zero and V is constant, equal to the surface value
The field inside a conductor vanishes in equilibrium, and since E = −dV/dr, zero field means the potential does not change — so V is constant throughout the cavity and equals its surface value. This is why a metal enclosure shields its contents, and why you are safe inside a car during a lightning strike.
Two capacitors of 2 μF and 3 μF are connected in series across a battery. Which is the same for both?
1. The potential difference across each
2. The charge on each
3. The energy stored in each
4. The capacitance of each
Show answer
Answer: B — The charge on each
In series the same charge must pass through the whole path, so Q is common and the voltages divide as V = Q/C — meaning the smaller capacitor takes the larger share of voltage. Parallel is the reverse: voltage is common and charges add. Note that this is the opposite convention from resistors, which is why the two are so often confused.
The electric field on the axis of a short dipole varies with distance as:
1. 1/r
2. 1/r²
3. 1/r³
4. Independent of r
Show answer
Answer: C — 1/r³
On the axis E = 2kp/r³. The extra power of r arises because the two opposite charges nearly cancel, and their small residual difference falls off faster than either field alone. The 1/r² answer is the single-point-charge law, offered because candidates apply it out of habit without noticing that a dipole has zero net charge.