Physics · Electrostatics · PYQ set
Electrostatics · JEE Main PYQ Practice with Solutions
Fifteen Electrostatics questions in the JEE Main style, from charge division and ring fields to capacitor networks and Gauss's law, each with a full worked solution and a concept recap.
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15Questions
JEE MainLevel
FullSolutions
In short
JEE Main returns to a small set of Electrostatics shapes: dividing a charge to maximise the force between the parts, the field on the axis of a ring, a dipole in a uniform field, choosing a Gaussian surface, and rearranging a capacitor network. This set works through fifteen of them with full solutions.
Contents
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Concept Recap
The formulae these past-year questions lean on most: fields and potentials of standard bodies, capacitor combinations and Gauss's law.
Q
Electrostatics – 15 JEE Main Questions
Single correct answer, in the JEE Main style. Attempt each before checking the key. Topics span the whole chapter.
Q1JEE MAINCoulomb's Law
A charge Q is divided into two parts q and (Q − q), placed a fixed distance apart. For the force between them to be maximum, q should be:
AQ/3
BQ/2
C2Q/3
DQ/4
Q2JEE MAINEquilibrium
Two charges each +q are at distance d. A third charge is placed at the midpoint. For the system to be in equilibrium, the third charge must be:
A+q/4
B−q
C−q/2
D−q/4
Q3JEE MAINContinuous Charge
The electric field at the centre of a uniformly charged ring of radius R is:
A2kQ/R²
BkQ/2R²
CZero
DkQ/R²
Q4JEE MAINContinuous Charge
The electric field on the axis of a charged ring is maximum at a distance from the centre of:
AR/2
BR/√2
C√2 R
DR
Q5JEE MAINConductors
A hollow conducting sphere of radius R carries charge Q. The potential at a distance r < R from the centre is:
AkQ/r²
BkQ/R
CZero
DkQ/r
Q6JEE MAINGauss's Law
Electric flux through a Gaussian surface enclosing charges +3q, −q and +2q is:
A4q/ε₀
B6q/ε₀
Czero
D2q/ε₀
Q7JEE MAINPotential
The work done to bring a unit positive charge from infinity to a point is 20 J. The potential at that point is:
A20 V
B20/ε₀ V
C−20 V
Dzero
Q8JEE MAINCapacitors
Two capacitors 2 μF and 3 μF are connected in series. The equivalent capacitance is:
A5 μF
B6 μF
C1.2 μF
D0.6 μF
Q9JEE MAINCapacitors
The same two capacitors 2 μF and 3 μF connected in parallel give:
A1.2 μF
B6 μF
C5 μF
D2.5 μF
Q10JEE MAINMotion in Field
A charged particle of mass m and charge q enters a uniform field E. Its acceleration is:
AqE/m
BqEm
CmE/q
Dq/mE
Q11JEE MAINDipole Energy
The potential energy of a dipole of moment p in a field E, when aligned with the field, is:
A−pE
B½pE
Czero
D+pE
Q12JEE MAINCombining Drops
Two identical drops of mercury each at potential V merge into one bigger drop. The potential of the big drop is:
A4V
BV
C2⁵∷³ V
D2V
Q13JEE MAINCapacitor Field
The electric field inside a parallel-plate capacitor with plate charge density σ is:
Aσ/2ε₀
Bσ/ε₀
C2σ/ε₀
Dzero
Q14JEE MAINApplication
A soap bubble is given a negative charge. Its radius:
AStays the same
BDecreases
CBecomes zero
DIncreases
Q15JEE MAINComparison
The ratio of electric force to gravitational force between two electrons is of the order:
A10²
B10²³
C10⁴⁰
D10⁴²
★
Answer Key & Solutions
Q1B
Q2D
Q3C
Q4B
Q5B
Q6A
Q7A
Q8C
Q9C
Q10A
Q11A
Q12C
Q13B
Q14D
Q15D
Q1Correct: BQ/2
Force ∝ q(Q−q). Maximising this product: d/dq[qQ − q²] = Q − 2q = 0, so q = Q/2. Splitting the charge equally maximises the force.
Q2Correct: D−q/4
For the outer charges to balance, the midpoint charge q' must satisfy kq·q'/(d/2)² = kq·q/d². Solving gives |q'| = q/4, and it must be negative to attract: −q/4.
Q3Correct: CZero
By symmetry, the field contributions from all elements of the ring cancel at the centre, giving zero. (The potential there is not zero, however.)
Q4Correct: BR/√2
The axial field E = kQx/(R²+x²)³∷² is maximised at x = R/√2, found by setting dE/dx = 0. A standard JEE Main result.
Q5Correct: BkQ/R
Inside a conductor the field is zero, so the potential is constant and equals its surface value kQ/R everywhere inside, not zero.
Q6Correct: A4q/ε₀
Only the net enclosed charge matters: 3q − q + 2q = 4q, so φ = 4q/ε₀. External charges contribute zero net flux.
Q7Correct: A20 V
Potential is defined as work done per unit charge to bring it from infinity: V = W/q = 20/1 = 20 V.
Q8Correct: C1.2 μF
In series, 1/C = 1/2 + 1/3 = 5/6, so C = 1.2 μF. Series capacitance is smaller than either capacitor.
Q9Correct: C5 μF
In parallel, capacitances add: C = 2 + 3 = 5 μF. Parallel capacitance is larger than either.
Q10Correct: AqE/m
Force F = qE, so acceleration a = F/m = qE/m. The particle accelerates along the field direction (for positive q).
Q11Correct: A−pE
U = −pE cosθ. At alignment θ = 0, so U = −pE, the minimum (most stable) energy. At θ = 180° it is +pE (unstable).
Q12Correct: C2⁵∷³ V
Volume doubles, so R' = 2¹∷³R and charge doubles. V' = kQ'/R' = k(2Q)/(2¹∷³R) = 2²∷³V = 2⁵∷³ V ≈ 1.59 V.
Q13Correct: Bσ/ε₀
Between the plates the two sheets' fields add: σ/2ε₀ + σ/2ε₀ = σ/ε₀. Outside, they cancel to zero.
Q14Correct: DIncreases
Like charges on the surface repel, creating an outward electrostatic pressure that expands the bubble, so the radius increases.
Q15Correct: D10⁴²
F_e/F_g = ke²/(Gm²) ≈ 10⁴². The electric force is overwhelmingly stronger, which is why gravity is negligible at the atomic scale.
★
Electrostatics: Quick Revision
Everything for revision day.MAX FORCE
split Q equally: Q/2
F ∝ q(Q−q)
peak at q = Q/2
RING
centre field: zero
axial max at x = R/√2
E = kQx/(R²+x²)³∷²
CONDUCTOR
V inside = kQ/R
field inside = 0
charge on surface
CAPACITORS
series: 1/C = Σ1/C_i
parallel: C = ΣC_i
U = ½CV²
GAUSS
φ = q_enclosed/ε₀
external charge: 0 net
cube face: q/6ε₀
DIPOLE ENERGY
U = −pE cosθ
aligned: −pE (stable)
torque = pE sinθ
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