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Electrostatic Potential and Capacitance is one of the biggest scoring chapters of JEE Main. Series and parallel combinations, energy stored, and the effect of a dielectric come up every single year. Before you start, fix one rule in your head: with the battery still connected the voltage is held, and with the battery removed the charge is held. Almost every trap in this chapter is built on that. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 2 to 3 questions every year, which is 8 to 12 marks.
- Capacitor combinations and energy stored are asked every year.
- Potential due to charge arrangements and dielectrics fill the rest.
What the syllabus says
In the syllabus (Unit 11, Electrostatics): Electric potential and its calculation for a point charge, a dipole and a system of charges, potential difference, equipotential surfaces, electrical potential energy of two point charges and of a dipole in an electrostatic field, conductors and insulators, dielectrics and electric polarization, capacitors and capacitance, capacitors in series and parallel, the parallel plate capacitor with and without a dielectric, and energy stored in a capacitor.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Potential of a point charge | V = kQ/r |
| Work done | W = q(VA − VB) |
| Field from potential | E = − dV/dr |
| Parallel plate capacitor | C = ε0A/d, and Kε0A/d with a dielectric |
| Isolated sphere | C = 4πε0R |
| Series and parallel | 1/C = 1/C1 + 1/C2, and C = C1 + C2 |
| Energy stored | U = ½CV2 = ½QV = Q2/2C |
| Energy density | u = ½ε0E2 |
| Common potential | V = (C1V1 + C2V2)/(C1 + C2) |
| Dipole potential energy | U = − pE cosθ |
Series capacitors behave the opposite way to series resistors: the reciprocals add, so the combination is always smaller than the smallest capacitor in it. Every capacitor in a series chain carries exactly the same charge.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. Capacitors of 2 µF and 4 µF are joined in series across a 12 V supply. The charge on each is
- (a) 8 µC
- (b) 16 µC
- (c) 24 µC
- (d) 48 µC
Q2. The energy stored in a 10 µF capacitor charged to 100 V is
- (a) 0.05 J
- (b) 0.5 J
- (c) 5 J
- (d) 0.005 J
Q3. When the gap of a parallel plate capacitor is completely filled with a dielectric of constant K, its capacitance
- (a) becomes K times larger
- (b) becomes K times smaller
- (c) stays the same
- (d) becomes K2 times larger
Q4. Two capacitors C and 2C are charged to the same potential V and then joined in parallel. Their common potential is
- (a) V
- (b) V/2
- (c) 2V
- (d) 3V
Q5. A charge q is placed at each corner of a square of side a. The potential at the centre of the square is
- (a) 4kq/a
- (b) 4√2 kq/a
- (c) zero
- (d) kq/a
Q6. The work done in carrying a charge between two points that are at the same potential is
- (a) qV
- (b) zero
- (c) q2V
- (d) infinite
Q7. The capacitance of an isolated spherical conductor of radius R is
- (a) 4πε0R
- (b) 4πε0R2
- (c) ε0R
- (d) 4πε0/R
Q8. A dielectric slab is pushed into a capacitor while the battery stays connected. The energy stored
- (a) becomes K times larger
- (b) becomes K times smaller
- (c) does not change
- (d) falls to zero
Q9. The equipotential surfaces around an isolated point charge are
- (a) parallel planes
- (b) concentric spheres
- (c) coaxial cylinders
- (d) ellipsoids
Q10. The potential energy of a dipole of moment p in a uniform field E, at an angle θ, is
- (a) pE sinθ
- (b) −pE cosθ
- (c) pE cosθ
- (d) pE
Answer key, with a hint for each
| Q1 | (b) | The series value is 4/3 µF, so Q = CV = (4/3) × 12. Both capacitors carry that same charge. |
| Q2 | (a) | U = ½ × 10 × 10−6 × 1002. |
| Q3 | (a) | C = Kε0A/d. The dielectric weakens the field inside and lets the plates hold more charge. |
| Q4 | (a) | They were already at the same potential, so no charge flows and nothing changes. |
| Q5 | (b) | Each charge is a√2/2 from the centre, so V = 4 × kq/(a/√2). |
| Q6 | (b) | Work depends on the potential difference, which is zero here. |
| Q7 | (a) | C = 4πε0R, so a bigger sphere holds more charge for the same potential. |
| Q8 | (a) | With the battery connected V is fixed, so U = ½CV2 rises with C. If the battery were removed the energy would fall instead. |
| Q9 | (b) | The potential depends only on r, so every point at the same distance is at the same potential. |
| Q10 | (b) | Lowest when the dipole lines up with the field, which is why the energy carries a minus and a cosine. |
Now mark yourself honestly
JEE Main marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 25 Physics questions are out of 100.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in January.
The full chapter, on a clock
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