Why this chapter carries the electricity block
Everything in Current Electricity, Magnetism and later electromagnetic induction is built on the field and potential ideas introduced here. Potential difference is what drives current; the capacitor work here reappears in AC circuits; and the field concept transfers wholesale to magnetism with a different source.
The chapter is also unusually practisable. Its questions divide fairly cleanly into two families - field and potential calculations for symmetric charge distributions, and capacitor network problems - and both respond well to repetition. That combination of high weight and high practisability makes it one of the best-value chapters in the paper.
How JEE actually asks Electrostatics
NTA does not publish chapter-wise weightage, so treat online figures as coaching estimates from past papers, and confirm the syllabus and pattern for your session in the official NTA information bulletin.
In past papers the questions split roughly into three groups. Direct applications of Coulomb's law and the field or potential of point charges are the easiest marks. Gauss's law questions appear for the three symmetric geometries and for conductors and shells. Capacitor questions - networks, energy, and the effect of dielectrics - are the most numerous single family, and the dielectric questions in particular reward knowing which quantity is held fixed.
Key concepts, compressed
- ✓Coulomb's law has the same inverse-square form as gravitation, but charges can repel as well as attract.
- ✓Electric field E is force per unit charge, a vector, pointing away from positive charge and toward negative.
- ✓Electric potential V is potential energy per unit charge, a scalar, and it adds arithmetically with signs.
- ✓The two are linked by E = −dV/dr: the field points down the steepest potential gradient.
- ✓In electrostatic equilibrium a conductor has zero field inside, all excess charge on its surface, and its entire volume at one potential.
- ✓Gauss's law states that the flux through any closed surface is the enclosed charge divided by ε₀, regardless of the surface's shape or of charges outside it.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Coulomb's law | F = kq₁q₂/r² | k = 9 × 10⁹ N·m²/C² |
| Field of a point charge | E = kq/r² | vector |
| Potential of a point charge | V = kq/r | scalar |
| Force on a charge in a field | F = qE | |
| Field from potential | E = −dV/dr | |
| Work moving a charge | W = qΔV | |
| Gauss's law | Φ = q(enclosed)/ε₀ | ε₀ = 8.85 × 10⁻¹² F/m |
| Field of an infinite sheet | E = σ/2ε₀ | independent of distance |
| Parallel-plate capacitance | C = ε₀A/d | ×K with a dielectric |
| Charge on a capacitor | Q = CV | |
| Energy stored | U = ½CV² = Q²/2C | |
| Capacitors in series | 1/C = 1/C₁ + 1/C₂ | opposite to resistors |
| Capacitors in parallel | C = C₁ + C₂ |
The five mistakes that cost the most marks
- ✓Adding electric fields arithmetically. E is a vector and must be added component-wise; only V adds as a plain sum.
- ✓Assuming E = 0 implies V = 0. Between two equal positive charges the field is zero at the midpoint while the potential is at a maximum there.
- ✓Using Gauss's law where there is no symmetry. It remains true, but you cannot take E outside the integral, so it yields nothing computable.
- ✓Reversing the series and parallel rules for capacitors. They are the opposite of resistors, and this is the single most common circuit error in the chapter.
- ✓Handling a dielectric without first asking whether the battery is connected. With it connected V is fixed and Q rises; disconnected, Q is fixed and V falls.
Practice set 1: Coulomb's law
1. Two charges of 1 μC each are placed 1 m apart in vacuum. What is the force between them?
9 × 10⁻³ N, repulsive. F = kq₁q₂/r² = 9 × 10⁹ × (10⁻⁶)(10⁻⁶)/1² = 9 × 10⁹ × 10⁻¹² = 9 × 10⁻³ N. Both charges are positive, so the force is repulsive. Note how small a coulomb-scale force this is - which tells you a microcoulomb is already a substantial charge.
2. The separation between two charges is halved. What happens to the force?
It becomes four times larger. The force goes as 1/r², so halving r multiplies F by four. This is identical in form to Newton's law of gravitation, and the same scaling arguments transfer directly between the two chapters.
3. Two charges are placed in a medium of dielectric constant 2 instead of vacuum. What happens to the force?
It halves. The force in a medium is F(vacuum)/K, where K is the dielectric constant, so with K = 2 the force is reduced to one half. Physically, the medium polarises and its induced charges partially screen the original ones - which is also why a dielectric increases capacitance.
4. Charge A exerts a force on charge B. A third charge C is brought nearby. Does the force between A and B change?
No. Electrostatic forces obey superposition: the force between any pair is unaffected by other charges present. What changes is the net force on each charge, which is now the vector sum of two contributions. Questions rely on students confusing 'the force from A' with 'the total force on B'.
5. Three equal positive charges sit at the vertices of an equilateral triangle. What is the direction of the net force on each?
Radially outward, away from the centroid, along the line from the centre through that vertex. By symmetry the two forces on any charge are equal in magnitude and make equal angles with that line, so their components perpendicular to it cancel. Symmetry arguments like this save far more time here than direct vector arithmetic.
6. Two point charges are separated by a distance r. If both charges are doubled and the distance is also doubled, what happens to the force?
It is unchanged. The numerator increases by a factor of four while the denominator also increases by four, so the ratio is the same. Scaling questions of this type are common because they test whether you can read the structure of the formula rather than substitute numbers.
Practice this now
Practice set 2: electric field
7. What is the electric field 0.1 m from a point charge of 1 μC?
9 × 10⁵ N/C, directed away from the charge. E = kq/r² = 9 × 10⁹ × 10⁻⁶/(0.1)² = 9 × 10³/10⁻² = 9 × 10⁵ N/C. The direction is radially outward because the charge is positive; for a negative charge the magnitude is the same but the field points inward.
8. What is the electric field inside a conductor in electrostatic equilibrium?
Zero, everywhere within the conducting material. If a field existed, free charges would move in response - and equilibrium means they are not moving. This is why excess charge sits entirely on the surface, and it is the principle behind electrostatic shielding and the Faraday cage.
9. How does the field of an infinite charged sheet vary with distance from it?
It does not vary at all: E = σ/2ε₀ is independent of distance. Moving away reduces the contribution of nearby charge but brings more of the sheet into play at a favourable angle, and the two effects cancel exactly. This uniform field is what makes the parallel-plate capacitor analysis so simple.
10. Can two electric field lines intersect?
No. The tangent to a field line gives the direction of E at that point, and a crossing would mean the field had two different directions simultaneously, which is impossible. The same argument applies to gravitational field lines and to magnetic field lines.
11. Two equal positive charges are separated by distance d. What is the field at the midpoint?
Zero. The two contributions are equal in magnitude and opposite in direction at that point, so they cancel as vectors. Note carefully that the potential there is not zero - it is the sum of two positive contributions and is in fact a maximum along the line. This pairing is the chapter's signature question.
12. What force acts on a 2 μC charge placed in a field of 10⁵ N/C?
0.2 N, along the field direction. F = qE = 2 × 10⁻⁶ × 10⁵ = 0.2 N. A negative charge of the same magnitude would experience the same force in the opposite direction. This relation is also the definition of the field: E is simply the force per unit positive test charge.
Practice set 3: electric potential
13. What is the electric potential 0.1 m from a point charge of 1 μC?
9 × 10⁴ V. V = kq/r = 9 × 10⁹ × 10⁻⁶/0.1 = 9 × 10³/0.1 = 9 × 10⁴ V. Note the distinction from question 7: potential falls as 1/r while field falls as 1/r², so at 0.1 m the numbers differ by a factor of 0.1 - the distance itself.
14. Two equal positive charges are separated by distance d. What is the potential at the midpoint?
Not zero - it is 4kq/d, the sum of two contributions of 2kq/d each. Potential is a scalar, so two positive charges reinforce rather than cancel. Compare directly with question 11, where the field at that same point was zero. Zero field with non-zero potential is entirely possible, and this configuration is the standard example.
15. What is the relationship between electric field and potential?
E = −dV/dr: the field is the negative gradient of the potential. The minus sign means the field points in the direction of decreasing potential, which is why a positive charge released from rest moves toward lower potential. A consequence worth remembering is that where the potential is constant, the field is zero.
16. How much work is done moving a charge between two points on the same equipotential surface?
Zero. Since W = qΔV and ΔV = 0 along an equipotential, no work is required regardless of the path taken. It follows that field lines are always perpendicular to equipotential surfaces - if they were not, there would be a field component along the surface and moving along it would require work.
17. How much work is done moving a 2 μC charge through a potential difference of 100 V?
2 × 10⁻⁴ J. W = qΔV = 2 × 10⁻⁶ × 100 = 2 × 10⁻⁴ J. The sign matters in context: moving a positive charge to a higher potential requires positive work from an external agent, while moving it to a lower potential means the field does the work.
18. A charge is moved along a closed loop in an electrostatic field. What is the total work done?
Zero. The electrostatic force is conservative, so the work depends only on the endpoints - and for a closed loop those coincide. This is the electrostatic analogue of gravity being conservative, and it is precisely what allows a potential to be defined at all.
Practice this now
Practice set 4: Gauss's law and conductors
19. What is the total electric flux through a closed surface enclosing a charge of 8.85 × 10⁻¹² C?
1 N·m²/C. By Gauss's law, Φ = q(enclosed)/ε₀ = 8.85 × 10⁻¹²/8.85 × 10⁻¹² = 1 N·m²/C. The charge was chosen to make the arithmetic clean, but the point is structural: the flux depends only on the enclosed charge and on nothing else about the situation.
20. Does the flux through a closed surface depend on the shape or size of the surface?
No. Gauss's law gives the flux as the enclosed charge divided by ε₀ regardless of whether the surface is a sphere, a cube or an irregular blob, and regardless of where inside it the charge sits. Only the total enclosed charge matters. The field at individual points on the surface does depend on all of that - but the total flux does not.
21. A charge is placed just outside a closed surface. What is the net flux through that surface?
Zero. Every field line from an external charge that enters the surface must also leave it, so the inward and outward contributions cancel exactly. The field at points on the surface is certainly not zero, which is the distinction the question is testing.
22. What is the field inside and outside a uniformly charged conducting spherical shell of total charge q?
Zero everywhere inside, and kq/r² outside - identical to a point charge q at the centre. Applying Gauss's law to a spherical surface inside the shell encloses no charge, giving zero field. Outside, the enclosed charge is the full q. Note that the potential inside is not zero: it is constant at kq/R, the surface value.
23. Where does the excess charge on a charged conductor reside, and why?
Entirely on its outer surface. Like charges repel and move as far apart as possible, and any charge remaining in the interior would create an internal field, contradicting electrostatic equilibrium. This is also why a hollow conductor screens its interior from external fields.
Practice set 5: capacitors
24. On what does the capacitance of a parallel-plate capacitor depend?
On geometry and the medium only: C = ε₀A/d, so on the plate area, their separation, and the dielectric constant of the material between them - multiplied by K if a dielectric is present. It does not depend on the charge stored or the voltage applied. Capacitance is a property of the device, not of its current state.
25. A 10 μF capacitor is charged to 100 V. What charge does it store?
10⁻³ C, or 1 mC. Q = CV = 10 × 10⁻⁶ × 100 = 10⁻³ C. Converting microfarads to farads before substituting is essential - leaving C as 10 is the most common arithmetic error in capacitor questions.
26. How much energy is stored in that capacitor?
0.05 J. U = ½CV² = ½ × 10 × 10⁻⁶ × 100² = ½ × 10⁻⁵ × 10⁴ = 0.05 J. You can equally use U = Q²/2C = (10⁻³)²/(2 × 10⁻⁵) = 10⁻⁶/(2 × 10⁻⁵) = 0.05 J. Choosing the form that matches the quantities you were given saves a step.
27. Two 10 μF capacitors are connected in series. What is the effective capacitance?
5 μF. For capacitors in series, 1/C = 1/10 + 1/10 = 2/10, so C = 5 μF. Series capacitors give less capacitance than either individual one - the opposite of resistors, where series increases resistance. Getting this backwards is the chapter's most common circuit error.
28. The same two capacitors are connected in parallel. What is the effective capacitance?
20 μF. In parallel, capacitances add directly: C = 10 + 10 = 20 μF. Physically, connecting plates in parallel is equivalent to increasing the total plate area, and C is proportional to area. That physical picture makes the rule hard to forget.
29. A dielectric is inserted into a capacitor while it remains connected to a battery. What happens to the voltage, charge and energy?
The voltage stays fixed at the battery value, the capacitance rises by the factor K, so the charge Q = CV rises by K, and the energy U = ½CV² also rises by K. The extra charge and energy are supplied by the battery. The key move is recognising that a connected battery fixes V, which then determines everything else.
30. The same capacitor is disconnected from the battery before the dielectric is inserted. Now what happens?
The charge is fixed, since it has nowhere to go. Capacitance still rises by K, so the voltage V = Q/C falls by K, and the energy U = Q²/2C also falls by K. The energy decreases because the dielectric is pulled into the gap by the field, so the field does work on it. Compare with question 29 - same insertion, opposite energy change, purely because of what the circuit holds fixed.
How to study this chapter efficiently
- ✓Label every quantity as vector or scalar before combining anything. Field and force are vectors; potential, potential energy and flux are scalars.
- ✓Learn the three configurations where field and potential behave differently - the midpoint between like charges, the midpoint between unlike charges, and the interior of a charged shell. Together they cover most conceptual questions.
- ✓For Gauss's law, check symmetry first. If the distribution is not spherical, cylindrical or planar, the law will not give you a number.
- ✓Write 'capacitors are the opposite of resistors' at the top of your notes and check every network against it.
- ✓For any dielectric question, first write down whether the battery is connected. Everything else follows from that one decision.
- ✓Do capacitor network practice specifically. It is the single largest question family in the chapter and it is almost entirely a redrawing skill.
Turn this into active practice
The vector-versus-scalar distinction is one that students report knowing and then fail to apply under time pressure. It is not fixed by rereading the definitions - it is fixed by meeting a question where the field is zero and the potential is not, getting it wrong, and remembering the shape of that mistake.
The JEE Electrostatics quiz on QUFF generates fresh questions across Coulomb's law, field and potential, Gauss's law and capacitors, marks them instantly and explains each answer. Do a mixed timed set, and for each error note whether it was a vector-scalar confusion, a symmetry misjudgement or a circuit-rule reversal.