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Electrostatics for JEE: Field, Potential, Capacitors and Current Electricity
Electrostatics for JEE covers electric charge and field, Gauss's law, electric potential and capacitors; current electricity then covers Ohm's law, cells, Kirchhoff's laws, the Wheatstone and metre bridge, and meters. Learn them as one block, in that order, because every circuit question rests on the ideas of potential and energy that electrostatics builds. Most lost marks come from sign errors and from not knowing which quantity stays constant, not from hard physics.
This guide is part of our JEE Physics tutor in Gurgaon series. For each topic it gives key ideas, common traps and verified worked examples, then compares JEE Main and JEE Advanced using the official 2026 syllabi.
Why electrostatics and current electricity matter for JEE
In the official JEE (Main) 2026 syllabus these are two units, Unit 11: Electrostatics and Unit 12: Current Electricity. Their ideas (field, potential, current, resistance) return in magnetism, induction and alternating current.
The exam format rewards accuracy. According to the JEE (Main) 2026 Information Bulletin, physics has 25 questions for 100 marks: 20 multiple-choice questions in Section A and 5 numerical-value questions in Section B. Both sections carry +4 for a correct answer and −1 for a wrong one, and numerical answers are entered as integers. One sign error therefore costs five marks, not four.
You can read the full topic list in the official JEE (Main) 2026 syllabus. Check the current version each year, because NTA can revise it.
Study order at a glance
This is the order we teach the JEE electricity chapters in. Each step uses the one before it.
| Step | Topic | Must be solid before moving on |
|---|---|---|
| 1 | Coulomb's law and electric field | Adding fields as vectors; field of a dipole; torque on a dipole |
| 2 | Gauss's law | Choosing a Gaussian surface; wire, sheet and shell results |
| 3 | Electric potential and energy | Potential as a scalar sum; E = −dV/dr; potential energy of a system |
| 4 | Capacitors | Series/parallel; dielectrics; energy; what stays constant |
| 5 | Current, resistance and power | Drift velocity, resistivity, temperature dependence, power formulas |
| 6 | Cells and internal resistance | Terminal voltage; cells in series and parallel |
| 7 | Kirchhoff's laws | Consistent sign convention; solving two-loop circuits |
| 8 | Bridges and meters | Balance condition; metre bridge; shunts and series resistors |
Charge, Coulomb's law and electric field
Coulomb's law gives the force between two point charges: F = kq1q2/r2, where k = 1/(4πε0) ≈ 9 × 109 N m2 C−2. The electric field at a point is the force per unit positive test charge, E = F/q0. With several charges, you add the individual forces or fields as vectors (the superposition principle).
Key ideas
- Field of a point charge: E = kq/r2, pointing away from a positive charge and towards a negative one.
- Superposition: find each field separately, then add them as vectors using components. Use symmetry to cancel components before calculating.
- Electric dipole: two equal and opposite charges ±q separated by 2a, with dipole moment p = q(2a) pointing from −q to +q. Far from the dipole, the field on the axis is 2kp/r3 and on the perpendicular bisector (equatorial line) it is kp/r3, directed opposite to p.
- Dipole in a uniform field: the net force is zero, the torque is τ = pE sin θ, and the potential energy is U = −pE cos θ.
- Continuous charge distributions: split the object into small charges dq and integrate. The standard results are the ring on its axis, E = kQx/(x2 + R2)3/2, and the line charge.
Traps
- Adding field magnitudes as numbers instead of as vectors.
- Using the far-field dipole formulas at points close to the dipole, where they do not apply.
- Drawing the dipole moment from + to −. It points from the negative charge to the positive charge.
Worked example: Charges of +4 µC and +1 µC are placed 30 cm apart. Where, on the line joining them, is the electric field zero?
Answer: 20 cm from the +4 µC charge (10 cm from the +1 µC charge). For two like charges, the fields point in opposite directions only between them, so the null point lies between the charges. Let it be x metres from the 4 µC charge. Setting the magnitudes equal: k(4 µC)/x2 = k(1 µC)/(0.3 − x)2. Taking square roots: 2/x = 1/(0.3 − x), so 0.6 − 2x = x and x = 0.2 m. Trap: the squared equation also gives x = 0.6 m, but that point is outside the pair, where both fields point the same way, so it is rejected.
Electric flux and Gauss's law
Gauss's law states that the total electric flux through any closed surface equals the charge enclosed divided by ε0: Φ = qenclosed/ε0. It is always true, but it finds fields only when symmetry keeps E constant over a well-chosen surface.
Key ideas
- Flux through a flat surface in a uniform field is Φ = EA cos θ, where θ is the angle between E and the normal to the surface.
- Infinitely long line charge (charge per unit length λ): E = λ/(2πε0r), using a cylindrical Gaussian surface.
- Infinite plane sheet (surface charge density σ): E = σ/(2ε0), the same at every distance.
- Thin spherical shell (charge Q, radius R): E = 0 inside, and E = kQ/r2 outside, as if all the charge were at the centre.
- Just outside a conductor: E = σ/ε0, perpendicular to the surface. This is twice the field of an isolated sheet.
Traps
- Thinking the field on a Gaussian surface comes only from the enclosed charge. The flux does; the field depends on all charges.
- Assuming that zero field inside a shell means zero potential inside it. The potential inside a shell is constant, and equal to its value at the surface.
- Applying the sheet formula σ/(2ε0) to a conductor surface, where the answer is σ/ε0.
Worked example: A thin spherical shell of radius 10 cm carries a charge of 2 nC. Find the electric field at 5 cm and at 20 cm from the centre, and the potential at the centre. (Take k = 9 × 109 N m2 C−2.)
Answer: 0 at 5 cm; 450 N/C at 20 cm; 180 V at the centre. At 5 cm the point is inside the shell, so a Gaussian sphere there encloses no charge and E = 0. At 20 cm, E = kQ/r2 = (9 × 109 × 2 × 10−9)/(0.2)2 = 18/0.04 = 450 N/C. Inside the shell the potential equals its surface value, V = kQ/R = 18/0.1 = 180 V. Trap: answering 0 V at the centre because the field there is zero.
Electric potential and potential energy
Electric potential at a point is the work done per unit positive charge in bringing a test charge from infinity to that point without acceleration. For a point charge, V = kq/r. Potential is a scalar, so you add potentials with their signs, which is quicker than adding fields.
Key ideas
- Link between field and potential: E = −dV/dr. The field points in the direction in which potential decreases fastest.
- Potential difference and work: the work done by an external agent in moving charge q from A to B slowly is W = q(VB − VA).
- Equipotential surfaces are always perpendicular to field lines, and no work is done moving a charge along one.
- Potential energy of two charges: U = kq1q2/r, with signs. For a system, add the term for every pair once.
- Dipole potential: V = kp cos θ/r2 far from the dipole. On the equatorial line, V = 0.
Traps
- Dropping the sign of a negative charge when adding potentials or finding potential energy.
- Assuming V = 0 means E = 0. At the midpoint of a dipole, V = 0 but E is not zero.
- Assuming E = 0 means V = 0. At the midpoint between two equal positive charges, E = 0 but V is not zero.
Worked example: Three charges of +1 µC each are placed at the corners of an equilateral triangle of side 10 cm. How much work is needed to assemble them from far apart?
Answer: 0.27 J. The work needed equals the potential energy of the system. There are three pairs, each separated by 0.1 m, so U = 3 × kq2/a = 3 × (9 × 109 × (10−6)2)/0.1 = 3 × 0.09 = 0.27 J.
From our tutors: when students mix up field and potential, we ask them to write two lines before any calculation: "field is a vector, so I need directions" or "potential is a scalar, so I need signs". It sounds basic, but it prevents many of the errors we see in electrostatics tests, especially when a question asks for both quantities at one point.
Conductors, dielectrics and capacitors
A capacitor stores charge and energy. Its capacitance is C = Q/V, which depends only on its shape, size and the material between the plates, not on the charge. For JEE electrostatics and capacitors, the most important idea is knowing which quantity stays constant when something changes: if the battery stays connected, V is constant; if it is disconnected first, Q is constant.
Key ideas
- Conductors in electrostatic equilibrium: the field inside is zero, the whole conductor is at one potential, and any excess charge sits on the surface.
- Parallel plate capacitor: C = ε0A/d in vacuum, and C = Kε0A/d when completely filled with a dielectric of dielectric constant K.
- Partly filled: a dielectric slab of thickness t gives C = ε0A/(d − t + t/K). A metal slab of thickness t (K effectively infinite) gives C = ε0A/(d − t).
- Combinations: in parallel, C = C1 + C2 + …; in series, 1/C = 1/C1 + 1/C2 + …. Capacitors in series carry equal charge if they started uncharged.
- Energy stored: U = Q2/(2C) = CV2/2 = QV/2. The energy density in the field is ε0E2/2.
- Inserting a dielectric: with the battery disconnected, C becomes KC, V falls to V/K and U falls to U/K. With the battery connected, Q rises to KQ and U rises to KU.
Traps
- Using the series and parallel formulas for capacitors the same way as for resistors. They are the other way round.
- Choosing the wrong constant quantity (Q or V) when a dielectric is inserted or plates are moved.
- Assuming energy is conserved when two charged capacitors are connected. It is not; some energy is lost as heat and radiation in the connecting wires.
Worked example: A parallel plate capacitor has capacitance C0 in air. A dielectric slab of dielectric constant 2 and thickness half the plate separation is inserted. Find the new capacitance.
Answer: 4C0/3. Using C = ε0A/(d − t + t/K) with t = d/2 and K = 2: the denominator is d − d/2 + d/4 = 3d/4. So C = ε0A/(3d/4) = (4/3)(ε0A/d) = 4C0/3. Check: an air gap of d/2 (2C0) in series with a dielectric layer of d/2 (4C0) gives 8C0/6 = 4C0/3.
Worked example: A 2 µF capacitor is charged to 100 V and disconnected from the battery. It is then connected across an uncharged 3 µF capacitor. Find the common potential and the energy lost.
Answer: 40 V; 6 mJ lost. Charge is conserved: the initial charge is 2 µF × 100 V = 200 µC, shared by a total capacitance of 5 µF, so V = 200/5 = 40 V. Initial energy = ½ × 2 × 10−6 × 1002 = 10 mJ. Final energy = ½ × 5 × 10−6 × 402 = 4 mJ. So 6 mJ is lost as heat and radiation.
Is electrostatics or circuits the chapter pulling down your child's JEE physics score? Book a free JEE Physics demo class in Gurgaon. The tutor will test field, potential and capacitors step by step and show you exactly where the gap is.
Book a Free JEE Physics Demo +91 92204 75088Current, drift velocity, resistance and power
Current electricity for JEE starts with the electron picture. Current is the rate of flow of charge, I = dq/dt. In a metal wire, free electrons drift slowly against the field, and the current is I = neAvd, where n is the number of free electrons per unit volume and vd is the drift velocity.
Key ideas
- Drift velocity and mobility: vd = eEτ/m, where τ is the relaxation time. Mobility is drift speed per unit field, μ = vd/E.
- Ohm's law: V = IR for ohmic conductors at constant temperature. Non-ohmic devices, such as diodes, have non-linear I–V graphs.
- Resistance and resistivity: R = ρL/A. Resistivity depends on the material and temperature, not the shape. Conductivity is σ = 1/ρ.
- Temperature dependence: for metals, approximately ρ = ρ0[1 + α(T − T0)], so resistance rises with temperature.
- Stretching a wire: the volume stays constant, so if the length becomes n times, the area becomes 1/n times and R becomes n2 times.
- Power: P = VI = I2R = V2/R. Choose the form that uses the quantity that is the same for the elements you are comparing.
Traps
- Confusing the tiny drift velocity with the speed of the electrical signal.
- Using P = V2/R for resistors in series, where the current, not the voltage, is common.
Worked example: A 10 Ω wire is stretched uniformly to three times its original length. What is its new resistance?
Answer: 90 Ω. The volume LA is constant, so tripling L makes the area A/3. Then R = ρL/A becomes ρ(3L)/(A/3) = 9ρL/A = 9 × 10 = 90 Ω.
Worked example: Bulbs rated 100 W, 220 V and 60 W, 220 V are connected in series across a 220 V supply. Which glows brighter?
Answer: the 60 W bulb. From R = V2/P, the 100 W bulb has R = 2202/100 = 484 Ω and the 60 W bulb has R = 2202/60 ≈ 807 Ω. In series the current is the same, so power I2R is larger in the larger resistance, the 60 W bulb. Trap: assuming the higher rated bulb always glows brighter.
Cells, EMF and internal resistance
A cell's EMF (E) is the energy it supplies per unit charge. Real cells have an internal resistance r, so the voltage across the terminals falls when current is drawn: V = E − Ir while discharging, and V = E + Ir while being charged.
Key ideas
- Current in a simple circuit: I = E/(R + r).
- Cells in series: the EMFs add (with signs if any cell is reversed) and the internal resistances add.
- n identical cells in parallel: the EMF stays E and the internal resistance becomes r/n.
- Maximum power transfer: the external resistor receives the most power when R = r, and that power is E2/(4r).
Traps
- Treating terminal voltage and EMF as the same when current flows.
- Forgetting that a reversed cell in a series string subtracts its EMF but still adds its internal resistance.
Worked example: A cell of EMF 2 V and internal resistance 0.5 Ω is connected to a 3.5 Ω resistor. Find the current, the terminal voltage and the maximum power the cell could deliver to any external resistor.
Answer: 0.5 A; 1.75 V; 2 W. I = E/(R + r) = 2/(3.5 + 0.5) = 0.5 A. Terminal voltage V = E − Ir = 2 − 0.5 × 0.5 = 1.75 V (check: IR = 0.5 × 3.5 = 1.75 V). Maximum external power occurs when R = r = 0.5 Ω, and equals E2/(4r) = 4/2 = 2 W.
Kirchhoff's laws
Kirchhoff's laws solve any DC circuit that cannot be reduced to simple series and parallel combinations. The junction rule says the currents entering a junction equal the currents leaving it (conservation of charge). The loop rule says the sum of potential changes around any closed loop is zero (conservation of energy).
Key ideas
- Assign currents first, in any direction you like. A negative answer simply means the actual current flows the other way.
- Loop rule signs: going through a resistor in the direction of current, the potential drops by IR; going from the negative to the positive terminal of a cell, it rises by E.
- Node method: when several branches meet at two points, write every branch current in terms of one unknown node potential. This is often faster than writing loops.
Traps
- Changing the sign convention halfway through a problem.
- Rejecting a negative current as "wrong" instead of reading it as a reversed direction.
Worked example: Three branches connect the same two points A and B (B at 0 V). Branch 1 has a 12 V cell in series with 2 Ω; branch 2 has a 6 V cell in series with 2 Ω; branch 3 is a 4 Ω resistor. Both cells have their positive terminals towards A. Find the current in each branch. (Treat the cells as ideal.)
Answer: 2.4 A out of the 12 V cell; 0.6 A into the 6 V cell (it is being charged); 1.8 A through the 4 Ω resistor. Let A be at potential V. The currents flowing towards A are (12 − V)/2 in branch 1 and (6 − V)/2 in branch 2; the 4 Ω resistor carries V/4 away from A. Junction rule at A: (12 − V)/2 + (6 − V)/2 = V/4. Multiplying by 4: 24 − 2V + 12 − 2V = V, so V = 36/5 = 7.2 V. Then branch 1 carries (12 − 7.2)/2 = 2.4 A, branch 2 carries (6 − 7.2)/2 = −0.6 A, and the resistor carries 7.2/4 = 1.8 A. Check: 2.4 − 0.6 = 1.8 A. The negative sign shows that 0.6 A actually flows into the positive terminal of the 6 V cell.
From our tutors: in Kirchhoff problems we ask students to mark the assumed current directions and the potential at each junction on the diagram before writing a single equation. Students who skip this tend to write a correct equation for the wrong loop. A final check, adding the potential changes around one loop, takes seconds and catches most sign errors.
Wheatstone bridge and metre bridge
A Wheatstone bridge is four resistors P, Q, R and S arranged in a diamond, with a galvanometer across the middle. It is balanced, with no current through the galvanometer, when P/Q = R/S. Both the Wheatstone bridge and the metre bridge are named in the JEE (Main) 2026 syllabus under current electricity.
Key ideas
- Balanced bridge: the two ends of the galvanometer are at the same potential, so the middle branch can be removed when finding equivalent resistance.
- Metre bridge: a practical Wheatstone bridge using a uniform 1 m wire. With a known resistance R in one gap and an unknown X in the other, and the balance point at length l from the end next to R, R/X = l/(100 − l), with l in cm.
- Accuracy is best when the balance point is near the middle of the wire.
- Resistivity experiment: the JEE Main 2026 experimental skills list includes finding the resistivity of a wire's material using a metre bridge.
Traps
- Measuring l from the wrong end of the wire, which swaps l and 100 − l.
- Removing the middle branch from a bridge that is not balanced.
Worked example: In a Wheatstone bridge, P = 2 Ω, Q = 4 Ω, R = 3 Ω, S = 6 Ω, and the galvanometer has resistance 5 Ω. Find the equivalent resistance between the two ends of the bridge.
Answer: 3.6 Ω. P/Q = 2/4 = 1/2 and R/S = 3/6 = 1/2, so the bridge is balanced and no current flows through the galvanometer. Remove it: the arms P + Q = 6 Ω and R + S = 9 Ω are in parallel, giving 6 × 9/(6 + 9) = 54/15 = 3.6 Ω. The galvanometer's 5 Ω does not matter.
Worked example: In a metre bridge, a 6 Ω resistor is in the left gap and an unknown resistor X in the right gap. The balance point is 40 cm from the left end. Find X.
Answer: 9 Ω. 6/X = 40/(100 − 40) = 40/60, so X = 6 × 60/40 = 9 Ω. Sense check: the balance point is nearer the smaller resistance.
What about the potentiometer? Neither the JEE Main nor the JEE Advanced 2026 syllabus names it, although older question banks contain many potentiometer questions. Check the current syllabus before spending time on them.
Galvanometer, ammeter and voltmeter
A moving coil galvanometer detects small currents. It is converted into an ammeter by connecting a small resistance (a shunt) in parallel, and into a voltmeter by connecting a large resistance in series. The JEE Main 2026 syllabus places this conversion in the magnetic effects unit.
Key ideas
- Ammeter: for a galvanometer of resistance G and full-scale current Ig, a range of I needs a shunt S = IgG/(I − Ig).
- Voltmeter: a range of V needs a series resistance R = V/Ig − G.
- Ideal meters: an ideal ammeter has zero resistance and an ideal voltmeter infinite resistance.
Traps
- Putting the shunt in series or the voltmeter resistance in parallel.
- Forgetting that a real voltmeter draws current and changes the reading it is measuring.
Worked example: A galvanometer of resistance 50 Ω gives full-scale deflection at 2 mA. What is needed to convert it into (a) a 0–1 A ammeter and (b) a 0–10 V voltmeter?
Answer: (a) a shunt of about 0.1 Ω in parallel; (b) 4950 Ω in series. (a) S = IgG/(I − Ig) = (0.002 × 50)/(1 − 0.002) = 0.1/0.998 ≈ 0.1002 Ω. (b) R = V/Ig − G = 10/0.002 − 50 = 5000 − 50 = 4950 Ω.
RC circuits (JEE Advanced)
The JEE (Advanced) 2026 syllabus lists RC circuits with d.c. sources; the JEE (Main) 2026 syllabus does not name them. Add them after capacitors and Kirchhoff's laws are secure.
- Charging: q = CE(1 − e−t/RC). The time constant is τ = RC.
- Discharging: q = Q0e−t/RC.
- Steady state: after a long time, a capacitor in a DC circuit carries no current and behaves like a break in the circuit. At the instant of switching on, an uncharged capacitor behaves like a plain wire.
Worked example: A 2 µF capacitor is charged through a 1 kΩ resistor from a 10 V battery. Find the time constant and the charge after one time constant.
Answer: 2 ms; about 12.6 µC. τ = RC = 1000 × 2 × 10−6 = 2 × 10−3 s. The final charge is CE = 20 µC, and after one time constant q = 20(1 − e−1) ≈ 20 × 0.632 ≈ 12.6 µC, that is about 63% of the final charge.
JEE Main vs JEE Advanced: what each syllabus lists
Both exams test the same core ideas, but the official syllabi name slightly different topics. The comparison below is based on the JEE (Main) 2026 syllabus and the syllabus annexure in the JEE (Advanced) 2026 Information Brochure. Always check the current year's documents.
| Topic | JEE Main 2026 syllabus | JEE Advanced 2026 syllabus |
|---|---|---|
| Charge and field | Coulomb's law; superposition; continuous charge distributions; dipole field and torque | Coulomb's law; electric field; field lines |
| Gauss's law | Flux; Gauss's law for a long straight wire, infinite plane sheet and thin spherical shell | Flux; Gauss's law in simple cases (the same three examples) |
| Potential and energy | Potential of point charge, dipole and systems; equipotentials; potential energy of charges and dipoles | Potential; potential energy of a system of point charges and of dipoles in a uniform field |
| Capacitors | Conductors, dielectrics and polarisation; combinations; parallel plate with and without dielectric; energy | Parallel plate capacitor with and without dielectrics; series and parallel; energy stored |
| Current and resistance | Drift velocity, mobility; Ohm's law; I–V characteristics; power; resistivity; temperature dependence | Electric current; Ohm's law; heating effect of current |
| Cells and circuits | Internal resistance, EMF, cells in series and parallel; Kirchhoff's laws; Wheatstone bridge; metre bridge | Resistances and cells in series and parallel; Kirchhoff's laws and simple applications |
| Meters | Galvanometer conversion to ammeter and voltmeter (magnetic effects unit) | Moving coil galvanometer, voltmeter, ammeter and their conversions |
| RC circuits | Not named | RC circuits with d.c. sources (electromagnetic induction section) |
| Experiments | Resistivity using a metre bridge; resistance using Ohm's law; galvanometer resistance and figure of merit by half deflection | Ohm's law using a voltmeter and ammeter; specific resistance using a metre bridge and post office box |
| Potentiometer | Not named | Not named |
Beyond the lists, style differs. JEE Main questions usually test one or two ideas at a time, and speed matters. JEE Advanced questions more often combine ideas, such as a capacitor circuit that needs Kirchhoff's laws and energy together; its brochure says negative marks may apply to some questions, with the scheme given on each paper.
A preparation plan for the JEE electricity chapters
This plan follows the study order above and fits alongside Class 12 school work.
- Check the mechanics base (vectors, work–energy, equilibrium) before electrostatics starts.
- Learn each topic from NCERT first, then a JEE problem book for speed and variety.
- Keep a one-page sheet per topic of results, conditions and "what stays constant" rules. Our JEE physics formula sheet collects them in one place.
- Solve previous year questions chapter by chapter right after each topic, with an error log of mistake types. Our JEE physics PYQ strategy explains the method.
- Draw every circuit again in your own neat form before solving.
- Mix the chapters in the final months: capacitors in circuits, conductors with Gauss's law, meters with Kirchhoff's laws.
Next come magnetism and induction. Our guide to modern physics for JEE covers another Class 12 block that many students find quicker to secure.
Students in Gurgaon and across Gurugram often have full school timetables in Class 12, so the plan has to fit around school tests and board preparation. Board and JEE physics overlap strongly here, so the work supports board marks too.
Frequently asked questions
Is electrostatics hard for JEE?
Electrostatics feels hard mainly when vectors and mechanics are weak, because it uses the same tools. Once those are solid, most JEE electrostatics questions apply a small set of results (point charge, dipole, wire, sheet, shell) carefully, with signs and directions.
Should I study electrostatics before current electricity?
Yes. Current electricity uses potential difference and energy throughout, so studying electrostatics first makes circuits make sense physically rather than as a list of rules.
Is the potentiometer in the JEE Main 2026 syllabus?
No. The JEE (Main) 2026 syllabus lists Kirchhoff's laws, the Wheatstone bridge and the metre bridge under current electricity, but it does not name the potentiometer, and neither does the JEE (Advanced) 2026 syllabus. Check the current syllabus each year, as NTA can revise it.
What are the most common mistakes in electrostatics and capacitors for JEE?
Adding fields as numbers instead of vectors, dropping the sign of a negative charge, assuming zero field means zero potential, and holding the wrong quantity (charge or voltage) constant when a dielectric is inserted.
How do I get faster at Kirchhoff's law questions?
Redraw the circuit, mark assumed currents and junction potentials, and use the node method when many branches meet at two points. Read negative answers as reversed directions.
Are numerical-value questions common from these chapters?
Physics in JEE Main has 5 numerical-value questions per paper, and circuit and capacitor problems suit that format. With no options to check against, the habit of verifying answers (for example, with a loop sum) matters more.
Can a home tutor help with the JEE electricity chapter?
Yes. A one-to-one tutor can see exactly where a student's reasoning breaks down (vectors, signs or circuit redrawing) and fix it before moving on. Our JEE Physics tutors in Gurgaon teach at home or online.
Want electrostatics and circuits taught in the right order, with every sign and step checked? Book a free JEE Physics demo class with Ajay Vatsyayan Classes, Saraswati kunj II, Wazirabad, Sector 52, Gurugram, Haryana 122003. Male and female tutors are available, at home or online.
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