The electric field at a point inside a uniformly charged conducting spherical shell is:

kq/r², with r measured from the centre

kq/R², with R the radius of the shell

A closed surface encloses charges of +3q and −q. Another charge +5q lies outside it. The total electric flux through the surface is:

Zero, since the charges are of opposite sign

The work done in moving a point charge along an equipotential surface is:

Equal to qV, where V is the potential of the surface

Equal to the change in the charge’s kinetic energy

Maximum, since the field is strongest along an equipotential

A parallel-plate capacitor stays connected to a battery while a dielectric slab of constant K is inserted between its plates. Which statement is correct?

The potential difference stays the same and the charge rises K times

The charge stays the same and the potential difference falls K times

The potential difference stays the same and the charge rises K times

Both the charge and the potential difference rise K times

Neither changes, because the slab only fills space

Electrostatics — Charges, Fields, Potential and Capacitors

Electrostatics is the largest single block in the CUET Physics paper and the most formula-driven, which makes it the block to secure first. Almost every question is one of four shapes: a field or force from a formula, a flux from Gauss's law, an equivalent capacitance, or what happens when a dielectric goes in. None of them needs more than two lines of working at this register. What they do need is the condition attached to each formula — the sheet formula is for an infinite sheet, the shell result is for points inside a shell, the dielectric answer depends on whether the battery is still connected — and that is where the marks actually move.

Force, Field and Gauss's Law

Coulomb's law is F = kq₁q₂/r² with k = 9 × 10⁹ N m²/C² and ε₀ = 8.85 × 10⁻¹² C²/N m², and in a medium of dielectric constant K the force falls to F/K. The field is force per unit charge, E = F/q, so a point charge gives E = kq/r² pointing away from a positive charge. Gauss's law then handles every symmetric distribution in one line: the flux through any closed surface is q_enclosed/ε₀, independent of where inside the charge sits and of any charge outside.

The field results CUET asks, with the condition each one needs
Distribution	Field	The condition to check
Point charge	`E = kq/r²`	Inverse square. Doubling r divides E by four, not two.
Infinite line of charge	`E = λ/(2πε₀r)`	Inverse FIRST power of r, which is the whole point of the case.
Infinite charged sheet	`E = σ/(2ε₀)`	Independent of distance. The 2 disappears for a conductor's surface, where E = σ/ε₀.
Charged conducting shell	Outside: `kq/r²`. Inside: zero.	Outside it behaves as a point charge at the centre; inside there is no field at all, whatever the shell's charge.
Electric dipole, `p = q × 2a`	On the axis: `2kp/r³`. On the equator: `kp/r³`, opposite in direction.	Inverse CUBE, and the axial field is twice the equatorial at the same r.

🧠 A dipole in a uniform field: torque yes, force no

The two charges are equal and opposite, so in a uniform field the forces cancel and the net force is zero while the couple survives: τ = pE sin θ, maximum at 90° and zero at 0° or 180°. Potential energy is U = −pE cos θ, so 0° is stable equilibrium and 180° is unstable. In a NON-uniform field the forces no longer cancel and there is a net force too — which is why the word "uniform" in the stem is doing work.

Potential, and Capacitors as the Paper Sets Them

Potential is V = kq/r for a point charge and it is a scalar, so potentials from several charges add with signs and no components — which is why a potential question is usually faster than a field question with the same charges. Work done in moving a charge is W = qΔV, so moving along an equipotential surface costs nothing, and equipotentials are always perpendicular to the field lines. Inside a conductor the field is zero and the whole conductor sits at one potential, with all its charge on the outer surface.

C = Q/V, and for a parallel-plate capacitor C = ε₀A/d — so capacitance depends on geometry alone, not on how much charge is on it.
In series, 1/C = 1/C₁ + 1/C₂ and the CHARGE is the same on each; in parallel, C = C₁ + C₂ and the VOLTAGE is the same across each. This is the reverse of resistors, which is the confusion the paper counts on.
Energy is U = ½CV² = ½QV = Q²/2C. Pick the form whose two quantities the question actually gives, rather than converting.
A dielectric of constant K multiplies capacitance: C = KC₀. Whether Q or V then changes depends on the battery, and that is the next note.

⚠️ The dielectric question turns on one word: connected

With the battery still connected, V is fixed by the battery, so C rises by K, Q = CV rises by K, and the energy ½CV² rises by K. With the battery disconnected, Q is trapped on the plates, so C rises by K, V = Q/C falls by K, and the energy Q²/2C falls by K. Same slab, opposite answers — read the stem for "connected to the battery" or "isolated" before computing anything.

Key takeaways

Field falls as 1/r² for a point charge, 1/r for an infinite line, and not at all with distance for an infinite sheet (σ/2ε₀). Inside a charged conducting shell the field is exactly zero.
Gauss: flux through a closed surface is q_enclosed/ε₀ — independent of where the charge sits inside and of every charge outside. A dipole field goes as 1/r³, and the axial field is twice the equatorial at the same r.
Capacitors are the reverse of resistors: series adds reciprocals with charge common, parallel adds directly with voltage common. Energy is ½CV² = ½QV = Q²/2C.
A dielectric multiplies C by K always. With the battery connected, Q and energy rise by K at constant V; with it disconnected, V and energy fall by K at constant Q.

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.

Capacitors of 3 µF and 6 µF are joined in series. The equivalent capacitance is:
1. 9 µF
2. 2 µF
3. 4.5 µF
4. 18 µF
Show answer
Answer: B — 2 µF
In series the reciprocals add: 1/C = 1/3 + 1/6 = 1/2, so C = 2 µF. Note that the result is smaller than either capacitor, which is the quick sanity check for a series combination. Answering 9 µF adds them directly, which is the parallel rule and the single commonest error in this topic because it is the resistor habit carried across. 4.5 µF is the arithmetic mean and corresponds to no rule. 18 µF is the product without the sum, half of the product-over-sum shortcut.
The electric field at a point inside a uniformly charged conducting spherical shell is:
1. kq/r², with r measured from the centre
2. Zero
3. kq/R², with R the radius of the shell
4. σ/ε₀, directed outward
Show answer
Answer: B — Zero
A Gaussian surface drawn anywhere inside the shell encloses no charge, so the flux and therefore the field are zero — all the charge sits on the shell itself. kq/r² is the field OUTSIDE, where the shell behaves as a point charge at the centre, and it is the answer given by candidates who apply the outside formula everywhere. kq/R² is the value just at the surface, not inside. And σ/ε₀ is the field immediately outside a conductor's surface, which is a different location again — this question is really asking whether the three regions are being kept apart.
A closed surface encloses charges of +3q and −q. Another charge +5q lies outside it. The total electric flux through the surface is:
1. 7q/ε₀
2. 2q/ε₀
3. 4q/ε₀
4. Zero, since the charges are of opposite sign
Show answer
Answer: B — 2q/ε₀
Gauss's law counts only the enclosed charge, algebraically: (+3q) + (−q) = 2q, so the flux is 2q/ε₀ and the external +5q contributes nothing — its field enters the surface on one side and leaves on the other. 7q/ε₀ includes the outside charge, which is the error this question is built to catch. 4q/ε₀ adds the magnitudes without their signs. And zero would need the enclosed charges to sum to zero, which +3q and −q do not.
The work done in moving a point charge along an equipotential surface is:
1. Equal to qV, where V is the potential of the surface
2. Zero
3. Equal to the change in the charge’s kinetic energy
4. Maximum, since the field is strongest along an equipotential
Show answer
Answer: B — Zero
Work is W = qΔV, and on an equipotential surface ΔV is zero by definition, so no work is done however long the path is. The qV option computes the work to bring the charge from infinity to that surface, which is a different journey and the most tempting wrong answer. Equating the work to a change in kinetic energy assumes the charge is free and that no other force acts, neither of which the question states. And the last option inverts the geometry: the field is always perpendicular to an equipotential, so there is no component of it along the path at all.
A 2 µF capacitor is charged to 100 V. The energy stored in it is:
1. 0.02 J
2. 0.01 J
3. 2 × 10⁻⁴ J
4. 1 J
Show answer
Answer: B — 0.01 J
U = ½CV² = ½ × 2 × 10⁻⁶ × (100)² = ½ × 2 × 10⁻⁶ × 10⁴ = 0.01 J. Dropping the factor of ½ gives 0.02 J, which is the commonest slip and the reason the option is placed first. 2 × 10⁻⁴ J is CV, the charge rather than an energy, and the units alone rule it out. And 1 J is off by two orders of magnitude, which usually comes from mishandling the µ prefix — 1 µF is 10⁻⁶ F, so a microfarad-and-volts problem lands in the hundredths of a joule and not in joules.
A parallel-plate capacitor stays connected to a battery while a dielectric slab of constant K is inserted between its plates. Which statement is correct?
1. The charge stays the same and the potential difference falls K times
2. The potential difference stays the same and the charge rises K times
3. Both the charge and the potential difference rise K times
4. Neither changes, because the slab only fills space
Show answer
Answer: B — The potential difference stays the same and the charge rises K times
While the battery is connected it holds V fixed, so with C rising to KC the charge Q = CV rises K times and the stored energy ½CV² also rises K times, the extra energy coming from the battery. The first option is the correct answer to the OTHER version of this question, where the battery has been disconnected and Q is trapped — the two answers are mirror images and the stem's word "connected" is what selects between them. Both rising together is impossible with a battery fixing V. And "neither changes" ignores that the slab polarises and reduces the effective field, which is what capacitance measures.
Two charges attract with a force F in vacuum. If the separation is halved and the space between them is filled with a medium of dielectric constant 2, the force becomes:
1. 8F
2. 2F
3. 4F
4. F/2
Show answer
Answer: B — 2F
Force goes as 1/(K r²), so the two changes act in opposite directions: halving r multiplies the force by 4, and a dielectric constant of 2 divides it by 2, giving 4F/2 = 2F. Answering 8F multiplies by 4 and then by 2 instead of dividing, which treats the medium as strengthening the force when a dielectric always weakens it. 4F applies the distance change alone and ignores the medium. And F/2 applies the medium alone and ignores the distance — each wrong option drops exactly one of the two effects.
An electric dipole is placed in a uniform electric field at 90° to the field. The net force and the torque on it are respectively:
1. 2qE and zero
2. Zero and pE
3. Zero and zero
4. qE and pE/2
Show answer
Answer: B — Zero and pE
In a uniform field the forces on the two equal and opposite charges cancel exactly, so the net force is zero, while the couple they form gives τ = pE sin 90° = pE, which is the maximum torque. The 2qE option adds the two forces instead of cancelling them, which would be right only in a non-uniform field where they differ. Zero torque belongs to 0° or 180°, where the dipole is aligned. And qE with pE/2 mixes a single charge's force with a half-angle that appears in no formula here.